19 bài tập Toán 9 Kết nối tri thức Bài 9. Biến đổi đơn giản và rút gọn biểu thức chứa căn thức bậc hai có đáp án

a) $\sqrt {96.125} = \sqrt {{2^5}{{.3.5}^3}} = \sqrt {{2^4}{{.5}^2}.30} = 20\sqrt {30}$.

b) $\sqrt {{a^4}{b^5}} = {a^2}{b^2}\sqrt b$.

c) $\sqrt {{a^6}{b^{11}}} = \left| {{a^3}.{b^5}} \right|\sqrt b$.

d) Với $a > 1:$ $\sqrt {{a^3}{{\left( {1 - a} \right)}^4}} = \sqrt {{a^2}.a.{{\left[ {{{\left( {1 - a} \right)}^2}} \right]}^2}} = a{\left( {1 - a} \right)^2}\sqrt a$.

a) Với $x \ge 0$: $x\sqrt {13} = \sqrt {13{x^2}}$.

b) Với $x < 0$:$x\sqrt 2 = - \sqrt {2{x^2}}$.

c) Với $x < 0$:$x\sqrt { - \frac{{11}}{x}} = - \sqrt { - \frac{{11}}{x}{x^2}} = - \sqrt { - 11x}$.

a) Ta có $\sqrt {50} - \sqrt {32} + 3\sqrt 8 = \sqrt {25.2} - \sqrt {16.2} + 3\sqrt {4.2} = 5\sqrt 2 - 4\sqrt 2 + 3.2.\sqrt 2 = 7\sqrt 2$.

b) $\sqrt {25a} + 2\sqrt {160a} - 3\sqrt {10a} = \sqrt {25.10a} + 2.\sqrt {16.10a} - 3\sqrt {10a}$

$= 5\sqrt {10a} + 2.4.\sqrt {10a} - 3\sqrt {10a} = 10\sqrt {10a}$.

c) $\left( {2\sqrt 7 + \sqrt 3 } \right)\sqrt 7 - \sqrt {84} = 2\sqrt 7 .\sqrt 7 + \sqrt 3 .\sqrt 7 - \sqrt {4.21}$

$= 2.7 + \sqrt {21} - 2\sqrt {21} = 14 - \sqrt {21}$.

d) $\left( {\sqrt {63} - \sqrt 8 - \sqrt 7 } \right)\sqrt 7 + 2\sqrt {14} = \sqrt {63} .\sqrt 7 - \sqrt 8 .\sqrt 7 - \sqrt 7 .\sqrt 7 + 2\sqrt {14}$

$= \sqrt {9.7.7} - 2\sqrt 2 .\sqrt 7 - 7 + 2\sqrt {14} = 3.7 - 2\sqrt {14} - 7 + 2\sqrt {14} = 14$.

a) $\left( {\sqrt {2x} - 1} \right)\left( {2x + \sqrt {2x} + 1} \right) = \left( {\sqrt {2x} - 1} \right)\left( {{{\left( {\sqrt {2x} } \right)}^2} + \sqrt {2x} .1 + {1^2}} \right) = {\left( {\sqrt {2x} } \right)^3} - 1 = 2x\sqrt 2 x - 1$.

b) $\left( {\sqrt x + 2\sqrt y } \right)\left( {x - 2\sqrt {xy} + 4y} \right) = \left( {\sqrt x + 2\sqrt y } \right)\left[ {{{\left( {\sqrt x } \right)}^2} - \sqrt x .2\sqrt y + {{\left( {2\sqrt y } \right)}^2}} \right]$

$= {\left( {\sqrt x } \right)^3} + {\left( {2\sqrt y } \right)^3} = x\sqrt x + 8y\sqrt y$.

c) $\left( {\sqrt x + \sqrt y } \right)\left( {2\sqrt x - \sqrt y } \right) = \sqrt x .2\sqrt x - \sqrt x .\sqrt y + \sqrt y .2\sqrt x - \sqrt y .\sqrt y$

$= 2x - \sqrt {xy} + 2\sqrt {xy} - y = 2x + \sqrt {xy} - y$.

a) Với $x > 0,\,y > 0$:

$\frac{{\left( {x\sqrt y - y\sqrt x } \right)\left( {2\sqrt y + 2\sqrt x } \right)}}{{2\sqrt {xy} }} = \frac{{\sqrt {xy} \left( {\sqrt x - \sqrt y } \right)2.\left( {\sqrt x + \sqrt y } \right)}}{{2\sqrt {xy} }}$$= \left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right) = x - y.$

Vậy $\frac{{\left( {x\sqrt y - y\sqrt x } \right)\left( {2\sqrt y + 2\sqrt x } \right)}}{{2\sqrt {xy} }} = x - y$ với $x > 0,\,y > 0$.

b) Với $x \ge 5$:

${\left( {\sqrt 5 + \sqrt {x - 5} } \right)^2} = {\left( {\sqrt 5 } \right)^2} + 2.\sqrt 5 .\sqrt {x - 5} + {\left( {\sqrt {x - 5} } \right)^2}$$= 5 + 2\sqrt {5x - 25} + x - 5 = x + 2\sqrt {5x - 25}$.

Vậy $x + 2\sqrt {5x - 25} = {\left( {\sqrt 5 + \sqrt {x - 5} } \right)^2}$ với $x \ge 5$.

Ta có: $2\sqrt {\frac{3}{{20}}} = 2\frac{{\sqrt {60} }}{{20}} = 2.\frac{{2\sqrt {15} }}{{20}} = \frac{{\sqrt {15} }}{5}$

$$\begin{array}{l}\sqrt {\frac{1}{{60}}} = \frac{{\sqrt {60} }}{{60}} = \frac{{2\sqrt {15} }}{{60}} = \frac{{\sqrt {15} }}{{30}}\\\sqrt {\frac{1}{{15}}} = \frac{{\sqrt {15} }}{{15}}\end{array}$$

Vậy $2\sqrt {\frac{3}{{20}}} + \sqrt {\frac{1}{{60}}} - \sqrt {\frac{1}{{15}}} = \frac{{\sqrt {15} }}{5} + \frac{{\sqrt {15} }}{{30}} - \frac{{\sqrt {15} }}{{15}} = \sqrt {15} \left( {\frac{1}{5} + \frac{1}{{30}} - \frac{1}{{15}}} \right) = \frac{{\sqrt {15} }}{6}$.

a) Ta có $\frac{9}{{\sqrt 3 }} = \frac{{9\sqrt 3 }}{3} = 3\sqrt 3$

b) $\frac{3}{{\sqrt 5 - \sqrt 2 }} = \frac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{5 - 2}} = \sqrt 5 + \sqrt 2$

c). $\frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}} = \frac{{{{\left( {\sqrt 2 + 1} \right)}^2}}}{{2 - 1}} = 3 + 2\sqrt 2$

d) $\frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}}}{{5 - 3}} = \frac{{8 - 2\sqrt {15} }}{2} = 4 - \sqrt {15}$

e) $\frac{{1 - a\sqrt a }}{{1 - \sqrt a }} = \frac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}{{1 - \sqrt a }} = 1 + \sqrt a + a$

f) $\frac{1}{{\sqrt {18} + \sqrt 8 - 2\sqrt 2 }} = \frac{1}{{3\sqrt 2 + 2\sqrt 2 - 2\sqrt 2 }} = \frac{1}{{3\sqrt 2 }} = \frac{{\sqrt 2 }}{6}$

g) $\frac{{\sqrt 2 }}{{1 + \sqrt 2 - \sqrt 3 }} = \frac{{\sqrt 2 \left( {1 + \sqrt 2 + \sqrt 3 } \right)}}{{3 + 2\sqrt 2 - 3}} = \frac{{1 + \sqrt 2 + \sqrt 3 }}{2}$

h) $\frac{1}{{\sqrt 3 + \sqrt 2 - \sqrt 5 }} = \frac{{\sqrt 3 + \sqrt 2 + \sqrt 5 }}{{5 + 2\sqrt 6 - 5}} = \frac{{\sqrt 3 + \sqrt 2 + \sqrt 5 }}{{2\sqrt 6 }} = \frac{{3\sqrt 2 + 2\sqrt 3 + \sqrt {30} }}{{12}}$.

a) $A = \frac{1}{{7 + 4\sqrt 3 }} + \frac{1}{{7 - 4\sqrt 3 }} = \frac{{7 - 4\sqrt 3 + 7 + 4\sqrt 3 }}{{49 - 48}} = 14$

b) $B = \frac{{15}}{{\sqrt 6 + 1}} + \frac{4}{{\sqrt 6 - 2}} - \frac{{12}}{{3 - \sqrt 6 }} - \sqrt 6 = \frac{{15\left( {\sqrt 6 - 1} \right)}}{{6 - 1}} + \frac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} - \frac{{12\left( {3 + \sqrt 6 } \right)}}{{9 - 6}} - \sqrt 6$.

$$= 3\left( {\sqrt 6 - 1} \right) + 2\left( {\sqrt 6 + 2} \right) - 4\left( {3 + \sqrt 6 } \right) - \sqrt 6 = - 11$$

a). Ta có $\frac{{\sqrt a }}{{\sqrt a - \sqrt b }} - \frac{{\sqrt b }}{{\sqrt a + \sqrt b }} - \frac{{2b}}{{a - b}}{\rm{ (a}} \ge {\rm{0}}{\rm{,b}} \ge {\rm{0}}{\rm{,a}} \ne {\rm{0);}}$

$$\begin{array}{l}{\rm{ = }}\frac{{\sqrt a \left( {\sqrt a + \sqrt b } \right) - \sqrt b \left( {\sqrt a - \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right).\left( {\sqrt a + \sqrt b } \right)}} - \frac{{2b}}{{a - b}}\\ = \frac{{a + \sqrt {ab} - \sqrt {ab} + b}}{{a - b}} - \frac{{2b}}{{a - b}}\end{array}$$

$$= \frac{{a - b}}{{a - b}} = 1$$

b). Ta có $\frac{{a\sqrt b + b}}{{a - b}}\sqrt {\frac{{ab + {b^2} - 2\sqrt {a{b^3}} }}{{a\left( {a + 2\sqrt b } \right) + b}}} \left( {\sqrt a + \sqrt b } \right){\rm{ }}\left( {a > b > 0} \right)$

$$\begin{array}{l} = \frac{{\sqrt b \left( {a + \sqrt b } \right)}}{{\left( {\sqrt a + \sqrt b } \right).\left( {\sqrt a - \sqrt b } \right)}}\sqrt {\frac{{b\left( {a + b - 2\sqrt {ab} } \right)}}{{{a^2} + 2a\sqrt b + b}}} \left( {\sqrt a + \sqrt b } \right)\\ = \frac{{\sqrt b \left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)}}.\sqrt {\frac{{b{{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{{{\left( {a + \sqrt b } \right)}^2}}}} \\ = \frac{{\sqrt b \left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)}}.\frac{{\sqrt b \left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)}} = b\end{array}$$

a) Điều kiện: $x \ge 1$

$$\begin{array}{l}\frac{1}{2}\sqrt {x - 1} - \frac{3}{2}\sqrt {9x - 9} + 24\sqrt {\frac{{x - 1}}{{64}}} = - 17{\rm{ }} \Leftrightarrow x = 290.\\\frac{1}{2}\sqrt {x - 1} - \frac{3}{2}\sqrt {9(x - 1)} + 24\sqrt {\frac{{x - 1}}{{{8^2}}}} = - 17{\rm{ }}\\\frac{1}{2}\sqrt {x - 1} - \frac{9}{2}\sqrt {x - 1} + \frac{{24}}{8}\sqrt {x - 1} = - 17 \Leftrightarrow - \sqrt {x - 1} = - 17 \Leftrightarrow \sqrt {x - 1} = 17\\x - 1 = 289\\x = 290.\end{array}$$

Đối chiếu với điều kiện ta được: $x = 290.$

b) (Điều kiện $x \ge 0$

$$\begin{array}{l}3x - 7\sqrt x + 4 = 0\\3x - 3\sqrt x - 4\sqrt x + 4 = 0\\3\sqrt x \left( {\sqrt x - 1} \right) - 4\left( {\sqrt x - 1} \right) = 0\\\left( {\sqrt x - 1} \right)\left( {3\sqrt x - 4} \right) = 0\end{array}$$

$3\sqrt x - 4 = 0$ hoặc $\sqrt x - 1 = 0$

$\sqrt x = \frac{4}{3}$ hoặc $\sqrt x = 1$

$x = \frac{{16}}{9}$ hoặc $x = 1$

Đối chiếu với điều kiện ta được: $x = \frac{{16}}{9};x = 1$.

c) Điều kiện $x \ge 0$

$$\begin{array}{l} - 5x + 7\sqrt x + 12 = 0\\ - 5x - 5\sqrt x + 12\sqrt x + 12 = 0\\ - 5\sqrt x .\left( {\sqrt x + 1} \right) + 12\left( {\sqrt x + 1} \right) = 0\\\left( {\sqrt x + 1} \right)\left( { - 5\sqrt x + 12} \right) = 0\end{array}$$

$\sqrt x + 1 = 0$ hoặc $- 5\sqrt x + 12 = 0$

$\sqrt x = - 1$ (vô nghiệm) hoặc $\sqrt x = \frac{{12}}{5}$

$$x = \frac{{144}}{{25}}.$$

Đối chiếu điều kiện ta được: $x = \frac{{144}}{{25}}.$

a) (ĐK${\rm{ a}} \ge 0$)

$$\begin{array}{l}A = \frac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}} - \frac{{2a + \sqrt a }}{{\sqrt a }} + 1\\\,\,\,\,\, = \frac{{\sqrt a \left( {{{\sqrt a }^3} + 1} \right)}}{{a - \sqrt a + 1}} - \frac{{\sqrt a (2\sqrt a + 1)}}{{\sqrt a }} + 1 = \frac{{\sqrt a \left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{a - \sqrt a + 1}} - (2\sqrt a + 1) + 1\\\,\,\,\,\, = a + \sqrt a - 2\sqrt a - 1 + 1 = a - \sqrt a \end{array}$$

b) Với ${\rm{ a}} \ge 1$ hay $\sqrt a \ge 1$ nên $\sqrt a - 1 \ge 0$ và $a \ge 1$, suy ra $a \ge 0$

Suy ra $A = a - \sqrt a = \sqrt a \left( {\sqrt a - 1} \right) \ge 0$

Khi đó $A \ge 0 \Rightarrow \left| A \right| = A$

Vậy $\left| A \right| = A$.

c) Tìm $a$ để $A = 2$

$$\begin{array}{l}a - \sqrt a = 2\\a - \sqrt a - 2 = 0\\a + \sqrt a - 2\sqrt a - 2 = 0\\\sqrt a \left( {\sqrt a + 1} \right) - 2\left( {\sqrt a + 1} \right) = 0\\\left( {\sqrt a + 1} \right)\left( {\sqrt a - 2} \right) = 0\end{array}$$

$\sqrt a + 1 = 0$ hoặc $\sqrt a - 2 = 0$

$\sqrt a = - 1$ (vô nghiệm) hoặc $\sqrt a = 2$.

$$a = 4$$

Đối chiếu điều kiện ta được $a = 4$ thì ${\rm{A = 2}}$.

d). $A = a - \sqrt a = a - 2\sqrt a \frac{1}{2} + \frac{1}{4} - \frac{1}{4} = {\left( {\sqrt a - \frac{1}{2}} \right)^2} - \frac{1}{4}$

Vì ${\left( {\sqrt a - \frac{1}{2}} \right)^2} \ge 0$ với mọi a.

Nên ${\left( {\sqrt a - \frac{1}{2}} \right)^2} - \frac{1}{4} \ge - \frac{1}{4}$ với mọi a.

Vậy giá trị nhỏ nhất của A bằng $- \frac{1}{4}$ khi và chỉ khi $\sqrt a - \frac{1}{2} = 0 \Leftrightarrow a = \frac{1}{4}$

a). Điều kiện $- 1 < a < 1$

$$\begin{array}{l}B = \left( {\frac{3}{{\sqrt {a + 1} }} + \sqrt {1 - a} } \right):\left( {\frac{3}{{\sqrt {1 - {a^2}} }} + 1} \right)\\\,\,\,\, = \left( {\frac{{3 + \sqrt {1 - a} .\sqrt {a + 1} }}{{\sqrt {a + 1} }}} \right):\left( {\frac{{3 + \sqrt {1 - {a^2}} }}{{\sqrt {1 - {a^2}} }}} \right) = \frac{{3 + \sqrt {1 - {a^2}} }}{{\sqrt {a + 1} }}.\frac{{\sqrt {a + 1} .\sqrt {1 - a} }}{{3 + \sqrt {1 - {a^2}} }} = \sqrt {1 - a} \end{array}$$

b). Với $a = \frac{{\sqrt 3 }}{{2 + \sqrt 3 }};$Khi đó:

$$B = \sqrt {1 - a} = \sqrt {1 - \frac{{\sqrt 3 }}{{2 + \sqrt 3 }}} = \sqrt {\frac{{2 + \sqrt 3 - \sqrt 3 }}{{2 + \sqrt 3 }}} = \sqrt {\frac{{2.\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}} = \sqrt {\frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{4 - 3}}} = \sqrt 3 + 1$$

c). Với $- 1 < a < 1$, ta có:${\rm{ }}\sqrt B > B$ hay $\sqrt B \left( {1 - {\rm{ }}\sqrt B } \right) > 0$

TH1: $\left\{ \begin{array}{l}{\rm{ }}\sqrt B > 0\\1 - {\rm{ }}\sqrt B > 0\end{array} \right.$ nên $\left\{ \begin{array}{l}B > 0\\{\rm{ }}\sqrt B < 1\end{array} \right.$, suy ra $0 < B < 1$.

TH2: $\left\{ \begin{array}{l}{\rm{ }}\sqrt B < 0\\1 - {\rm{ }}\sqrt B < 0\end{array} \right.$ (vô nghiệm).

Khi đó $0 < \sqrt {1 - a} < 1$ hay $0 < 1 - a < 1$ nên $0 < a < 1.$

Kết hớp với điều kiện $- 1 < a < 1$ ta được $0 < a < 1$

Vậy $0 < a < 1$ thì${\rm{ }}\sqrt B > B$.

a). Với $a \ge 0$, ta có: $3\sqrt {2a} - \sqrt {18{a^3}} + 4\sqrt {\frac{a}{2}} - \frac{1}{4}\sqrt {128a}$

$= 3\sqrt {2a} - 3a\sqrt {2a} + 2\sqrt {2a} - 2\sqrt {2a}$

$= 3\sqrt {2a} - 3a\sqrt {2a} = 3\left( {1 - a} \right)\sqrt {2a} .$

b). Với $x > y > 0$, ta có: $2y\sqrt {x - y} + x\sqrt {\frac{1}{{x - y}}} - x\sqrt {\frac{a}{{ax - ay}}} - \sqrt {{x^3} - {x^2}y}$

$= 2y\sqrt {x - y} + x\sqrt {\frac{1}{{x - y}}} - x\sqrt {\frac{a}{{a\left( {x - y} \right)}}} - \sqrt {{x^2}\left( {x - y} \right)}$

$= 2y\sqrt {x - y} - \left| x \right|\sqrt {x - y} + x\sqrt {\frac{1}{{x - y}}} - x\sqrt {\frac{1}{{x - y}}}$

$= \sqrt {x - y} \left( {2y - x} \right)$ (do $x > 0$).

c). Với $a \ge 0,b \ge 0,a \ne b$, ta có:
$\frac{{\sqrt a + \sqrt b }}{{\sqrt a - \sqrt b }} + \frac{{\sqrt a - \sqrt b }}{{\sqrt a + \sqrt b }} = \frac{{{{\left( {\sqrt a + \sqrt b } \right)}^2} + {{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}$

$= \frac{{a + 2\sqrt {ab} + b + a - 2\sqrt {ab} + b}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}} = \frac{{2a + 2b}}{{a - b}}$.

a). Ta có: $\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 2}} - \frac{2}{{2 + \sqrt 2 }} + \frac{{\sqrt 2 + 1}}{{\sqrt 2 }} = \frac{{\sqrt 2 - 1}}{{\sqrt 2 (1 + \sqrt 2 )}} - \frac{2}{{\sqrt 2 (\sqrt 2 + 1)}} + \frac{{{{\left( {\sqrt 2 + 1} \right)}^2}}}{{\sqrt 2 (\sqrt 2 + 1)}}$

$= \frac{{\sqrt 2 - 1 - 2 + 2 + 2\sqrt 2 + 1}}{{\sqrt 2 (\sqrt 2 + 1)}} = \frac{{3\sqrt 2 }}{{\sqrt 2 (\sqrt 2 + 1)}} = \frac{3}{{\sqrt 2 + 1}} = \frac{{3\left( {\sqrt 2 - 1} \right)}}{{2 - 1}} = 3\left( {\sqrt 2 - 1} \right)$

b). Ta có: $\frac{{2 + \sqrt 5 }}{{\sqrt 2 + \sqrt {3 + \sqrt 5 } }} + \frac{{2 - \sqrt 5 }}{{\sqrt 2 - \sqrt {3 - \sqrt 5 } }} = \frac{{2\sqrt 2 + \sqrt {10} }}{{2 + \sqrt {6 + 2\sqrt 5 } }} + \frac{{2\sqrt 2 - \sqrt {10} }}{{2 - \sqrt {6 - 2\sqrt 5 } }}$

$= \frac{{2\sqrt 2 + \sqrt {10} }}{{2 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} + \frac{{2\sqrt 2 - \sqrt {10} }}{{2 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}$ $= \frac{{2\sqrt 2 + \sqrt {10} }}{{2 + \sqrt 5 + 1}} + \frac{{2\sqrt 2 - \sqrt {10} }}{{2 - \sqrt 5 + 1}} = \frac{{2\sqrt 2 + \sqrt {10} }}{{3 + \sqrt 5 }} + \frac{{2\sqrt 2 - \sqrt {10} }}{{3 - \sqrt 5 }}$

$= \frac{{\left( {2\sqrt 2 + \sqrt {10} } \right)\left( {3 - \sqrt 5 } \right) + \left( {2\sqrt 2 - \sqrt {10} } \right)\left( {3 + \sqrt 5 } \right)}}{{\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)}}$

$= \frac{{6\sqrt 2 - 2\sqrt {10} + 3\sqrt {10} - \sqrt {50} + 6\sqrt 2 + 2\sqrt {10} - 3\sqrt {10} - \sqrt {50} }}{{9 - 5}}$

$= \frac{{12\sqrt 2 - 2\sqrt {50} }}{4} = \frac{{12\sqrt 2 - 10\sqrt 2 }}{4} = \frac{{2\sqrt 2 }}{4} = \frac{{\sqrt 2 }}{2}.$

a) ${\rm{VT}} = \frac{{{{\left( {\sqrt {{a^2} + {x^2}} + \sqrt {{a^2} - {x^2}} } \right)}^2}}}{{{a^2} + {x^2} - \left( {{a^2} - {x^2}} \right)}} - \sqrt {\frac{{{a^4} - {x^4}}}{{{x^4}}}}$$= \frac{{2{a^2} + 2\sqrt {{a^4} - {x^4}} }}{{2{x^2}}} - \frac{{\sqrt {{a^4} - {x^4}} }}{{{x^2}}} = \frac{{{a^2}}}{{{x^2}}}$

b) Chú ý: $5 + 2\sqrt 6 = {(\sqrt 3 + \sqrt 2 )^2}$.

c) $x\sqrt x - y\sqrt y = {(\sqrt x )^3} - {(\sqrt y )^2} = (\sqrt x - \sqrt y )(x + \sqrt {xy} + y)$.

a) Với $x \ge 0$ và $x \ne 4$, ta có:$A = \frac{{\sqrt x + 1}}{{\sqrt x - 2}} + \frac{{2\sqrt x }}{{\sqrt x + 2}} + \frac{{2 + 5\sqrt x }}{{4 - x}}$

$= \frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{2\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} - \frac{{2 + 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}$

$= \frac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}$$= \frac{{3x - 6\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} = \frac{{3\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} = \frac{{3\sqrt x }}{{\sqrt x + 2}}$

b) Khi $A = 2$ ta được $\frac{{3\sqrt x }}{{\sqrt x + 2}} = 2 \Leftrightarrow 3\sqrt x = 2\left( {\sqrt x + 2} \right) \Leftrightarrow \sqrt x = 4 \Leftrightarrow x = 16{\rm{ }}(tm{\rm{ }}x \ge 0,x \ne 4)$

Vậy $x = 16$.

a) Với điều kiện: $a > b > 0$, ta có:$B = \frac{a}{{\sqrt {{a^2} - {b^2}} }} - \left( {1 + \frac{a}{{\sqrt {{a^2} - {b^2}} }}} \right):\frac{b}{{a - \sqrt {{a^2} - {b^2}} }}$

$\begin{array}{l} = \frac{a}{{\sqrt {{a^2} - {b^2}} }} - \frac{{\left( {\sqrt {{a^2} - {b^2}} + a} \right)}}{{\sqrt {{a^2} - {b^2}} }}.\frac{{\left( {a - \sqrt {{a^2} - {b^2}} } \right)}}{b}\\ = \frac{a}{{\sqrt {{a^2} - {b^2}} }} - \frac{b}{{\sqrt {{a^2} - {b^2}} }} = \frac{{\sqrt {a - b} }}{{\sqrt {a + b} }} = \sqrt {\frac{{a - b}}{{a + b}}} \end{array}$

b) Ta có: $\frac{a}{b} = \frac{3}{2}$$\Leftrightarrow a = \frac{3}{2}b$ thay vào ta được $B = \sqrt {\frac{{\frac{3}{2}b - b}}{{\frac{3}{2}b + b}}} = \sqrt {\frac{{\frac{1}{2}b}}{{\frac{5}{2}b}}} = \sqrt {\frac{1}{5}} = \frac{{\sqrt 5 }}{5}.$

c) Khi $B < 1$ ta có: $\sqrt {\frac{{a - b}}{{a + b}}} < 1 \Leftrightarrow$ $\frac{{\sqrt {a - b} }}{{\sqrt {a + b} }} < 1 \Leftrightarrow \sqrt {a - b} < \sqrt {a + b}$

$\Leftrightarrow a - b < a + b \Leftrightarrow - b < b \Leftrightarrow 2b > 0 \Leftrightarrow b > 0.$

Vậy với $a > b > 0$ thì $B < 1.$

a) ĐK: $x \ge 0,x \ne 1$, ta có:

$C = \left( {\frac{{\sqrt x - 2}}{{x - 1}} - \frac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\frac{{{{\left( {1 - x} \right)}^2}}}{2} = \left( {\frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}} \right).\frac{{{{\left( {1 - x} \right)}^2}}}{2}$

$= \frac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\frac{{{{\left( {1 - x} \right)}^2}}}{2} = \frac{{ - 2\sqrt x }}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\frac{{{{\left( {1 - x} \right)}^2}}}{2}$

$= \frac{{ - \sqrt x {{\left( {1 - x} \right)}^2}}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}} = \frac{{\sqrt x \left( {1 - x} \right)}}{{\left( {\sqrt x + 1} \right)}} = \sqrt x \left( {1 - \sqrt x } \right)$

b) Ta có:$C > 0 \Leftrightarrow \sqrt x \left( {1 - \sqrt x } \right) > 0 \Leftrightarrow 1 - \sqrt x > 0 \Leftrightarrow 1 > \sqrt x \Rightarrow 0 \le x < 1.$

c) Với $x \ge 0,x \ne 1$, ta có$C = \sqrt x \left( {1 - \sqrt x } \right) = \sqrt x - x = - \left( {x - \sqrt x } \right) = - \left( {x - 2\sqrt x .\frac{1}{2} + \frac{1}{4} - \frac{1}{4}} \right) = - {\left( {\sqrt x - \frac{1}{2}} \right)^2} + \frac{1}{4}$

Vì $- {\left( {\sqrt x - \frac{1}{2}} \right)^2} \le 0$ với mọi $x \ge 0$ nên $- {\left( {\sqrt x - \frac{1}{2}} \right)^2} + \frac{1}{4} \le \frac{1}{4}$ với mọi $x \ge 0$.

Do đó: $C \le \frac{1}{4}$ với mọi $x \ge 0$

GTLN của $C = \frac{1}{4}$ khi $\sqrt x - \frac{1}{2} = 0 \Leftrightarrow \sqrt x = \frac{1}{2} \Leftrightarrow x = \frac{1}{4}.$

Vậy GTLN của $C = \frac{1}{4}$ khi $x = \frac{1}{4}$.

Ta có: $\left( {\frac{{2\sqrt {xy} }}{{x - y}} + \frac{{\sqrt x - \sqrt y }}{{2\left( {\sqrt x + \sqrt y } \right)}}} \right).\frac{{2\sqrt x }}{{\sqrt x + \sqrt y }} + \frac{{\sqrt y }}{{\sqrt y - \sqrt x }}$

$$\begin{array}{l} = \frac{{4\sqrt {xy} + {{\left( {\sqrt x - \sqrt y } \right)}^2}}}{{2\left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x - \sqrt y } \right)}}.\frac{{2\sqrt x }}{{\sqrt x + \sqrt y }} + \frac{{\sqrt y }}{{\sqrt y - \sqrt x }}\\ = \frac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{2\left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x - \sqrt y } \right)}}.\frac{{2\sqrt x }}{{\sqrt x + \sqrt y }} + \frac{{\sqrt y }}{{\sqrt y - \sqrt x }}\\ = \frac{{\sqrt x }}{{\sqrt x - \sqrt y }} + \frac{{\sqrt y }}{{\sqrt y - \sqrt x }} = \frac{{\sqrt x }}{{\sqrt x - \sqrt y }} - \frac{{\sqrt y }}{{\sqrt x - \sqrt y }}\end{array}$$

$= \frac{{\sqrt x - \sqrt y }}{{\sqrt x - \sqrt y }} = 1$ (đpcm).

Vậy biểu thức đã cho không phụ thuộc vào biến.