15 bài tập Toán 9 Kết nối tri thức Bài 8. Khai căn bậc hai với phép nhân và phép chia có đáp án

Lời giải

a) $18;\,\,\,\,\,\,385$;

b) $5 - 2\sqrt 6 ;\,\,\,\,\,\,17;\,\,\,\,\,\sqrt 2 \left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right) = \sqrt 2$;

c) $2;\,\,\,\,\,\,\frac{1}{6}$

a) $\frac{5}{2};\,\,1\,;$ b) $11 + 6\sqrt 2 = {\left( {\sqrt 2 + 3} \right)^2};\,\,\,\,\sqrt {3 - \sqrt 3 }$

c) $- \frac{{21}}{2}c(a - b){b^3}.$ d) $12 - 18\sqrt 2 + 16\sqrt 3$

e) $ab + 2b - a + 1;$ g) ${a^3}bn - {a^3}{b^3} + {a^4}$

a) $\frac{{\sqrt 3 }}{{\sqrt 7 }};$ b) $\frac{{\sqrt 5 }}{2};$ c) $\frac{{\sqrt x }}{{\sqrt y }};$ d) $\frac{{\sqrt a - \sqrt b }}{{\sqrt {ab - 1} }};$ e) $\frac{{\sqrt 3 - \sqrt 2 }}{{1 - \sqrt 2 }}$

f) Tách $\sqrt {16} = 4 = \sqrt 4 + \sqrt 4$

$$\left( {\sqrt 2 + \sqrt 3 + \sqrt 4 } \right) + \left( {\sqrt 4 + \sqrt 6 + \sqrt 8 } \right) = \left( {\sqrt 2 + \sqrt 3 + \sqrt 4 } \right)\left( {1 + \sqrt 2 } \right)$$

Kết quả: $1 + \sqrt 2$.

$\begin{array}{l}\frac{1}{{\sqrt x }} + \frac{1}{{\sqrt y }} = 4 - \sqrt x - \sqrt y \\\left( {\sqrt x + \frac{1}{{\sqrt x }} - 2} \right) + \left( {\sqrt y + \frac{1}{{\sqrt y }} - 2} \right) = 0\end{array}$

$\begin{array}{l}\frac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} + \frac{{{{\left( {\sqrt y - 1} \right)}^2}}}{{\sqrt y }} = 0\\x - y = 1\end{array}$

a) $\frac{{a + b}}{2} - \sqrt {ab} = \frac{1}{2}{\left( {\sqrt a - \sqrt b } \right)^2} \ge 0$

b) $a + b + c - \sqrt {ab} - \sqrt {bc} - \sqrt {ca} = \frac{1}{2}\left[ {{{\left( {\sqrt a - \sqrt b } \right)}^2} + {{\left( {\sqrt b - \sqrt c } \right)}^2} + {{\left( {\sqrt c - \sqrt a } \right)}^2}} \right] \ge 0\,;$

c) $a + b + \frac{1}{2} - \sqrt a - \sqrt b = {\left( {\sqrt 1 - \frac{1}{2}} \right)^2} + {\left( {\sqrt b - \frac{1}{2}} \right)^2} \ge 0$.

a) Ta có$\,\sqrt {2\frac{7}{{81}}} = \sqrt {\frac{{169}}{{81}}} = \frac{{\sqrt {169} }}{{\sqrt {81} }} = \frac{{13}}{9}.$ và $\,\frac{{\sqrt 6 }}{{\sqrt {150} }} = \sqrt {\frac{6}{{150}}} = \sqrt {\frac{1}{{25}}} = \frac{1}{5}.$

b) Ta có $\,\left( {5\sqrt 7 + 7\sqrt 5 } \right):\sqrt {35} = \frac{{5\sqrt 7 }}{{\sqrt {35} }}\,\,\,\, + \frac{{7\sqrt 5 }}{{\sqrt {35} }} = \frac{5}{{\sqrt 5 }}\, + \frac{7}{{\sqrt 7 }} = \sqrt 5 \, + \sqrt 7 .$

c) Ta có $\,\left( {2\sqrt 8 - 3\sqrt 3 + 1} \right):\sqrt 6 = \frac{{2\sqrt 8 }}{{\sqrt 6 }} - \frac{{3\sqrt 3 }}{{\sqrt 6 }} + \frac{1}{{\sqrt 6 }} = \frac{{4\sqrt 3 }}{3} - \frac{{3\sqrt 2 }}{2} + \frac{{\sqrt 6 }}{6}.$

Tính $\frac{1}{2}\sqrt {\frac{1}{2}} - \frac{3}{2}\sqrt {4{,}5} + \frac{2}{5}\sqrt {50}$

$$= \frac{1}{2}\sqrt {\frac{1}{2}} - \frac{3}{2}\sqrt {\frac{9}{2}} + \frac{2}{5}\sqrt {25.2} = \frac{1}{{2\sqrt 2 }} - \frac{9}{{2\sqrt 2 }} + 2\sqrt 2 = \frac{{1 - 9 + {{\left( {2\sqrt 2 } \right)}^2}}}{{2\sqrt 2 }} = 0$$

Vậy $\left( {\frac{1}{2}\sqrt {\frac{1}{2}} - \frac{3}{2}\sqrt {4{,}5} + \frac{2}{5}\sqrt {50} } \right):\frac{4}{{15}}\sqrt {\frac{1}{8}} = 0$

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

Ta có $\,\frac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2} = \frac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - \frac{{\left( {x - 2\sqrt {xy} + y} \right)\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x + \sqrt y }}$

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

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

c)$\,\frac{{x - 1}}{{\sqrt y - 1}}.\sqrt {\frac{{{{\left( {y - 2\sqrt y + 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^4}}}} = \frac{{x - 1}}{{\sqrt y - 1}}.\sqrt {\frac{{{{\left( {\sqrt y - 1} \right)}^4}}}{{{{\left( {x - 1} \right)}^4}}}} = \frac{{x - 1}}{{\sqrt y - 1}}.\frac{{{{\left( {\sqrt y - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{\sqrt y - 1}}{{x - 1}}.$

a) Với $a > b > 0$, ta có

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

Ta thấy $a = 36;\,b = 25$ thỏa mãn điều kiện.

Thay $a = 36;\,b = 25$ vào (1) ta có $A = \sqrt {\sqrt {36} - \sqrt {25} } = \sqrt 1 = 1$.

b) Với $x > 3$ ta có;

$B = \frac{{\sqrt {x - 3} }}{{\sqrt {\sqrt x + \sqrt 3 } }}:\frac{{\sqrt {\sqrt x - \sqrt 3 } }}{{\sqrt x }} = \sqrt {\frac{{x - 3}}{{\sqrt x + \sqrt 3 }}} :\sqrt {\frac{{\sqrt x - \sqrt 3 }}{{\sqrt x }}}$

$= \sqrt {\frac{{(\sqrt x - \sqrt 3 )(\sqrt x + \sqrt 3 )}}{{\sqrt x + \sqrt 3 }}.\frac{x}{{\sqrt x - \sqrt 3 }}} = \sqrt x \,\,\,(1)$

Ta thấy $x = 81$thỏa mãn điều kiện.

Thay $x = 81$ vào $\left( 1 \right)$ta có $B = \sqrt {81} = 9$.

c) Với $x < 4$ ta có

$\begin{array}{l}C = \sqrt {\frac{{{{\left( {x - 5} \right)}^4}}}{{{{\left( {4 - x} \right)}^2}}}} - \frac{{{x^2} - 25}}{{x - 4}} = \frac{{\sqrt {{{\left( {x - 5} \right)}^4}} }}{{\sqrt {{{\left( {4 - x} \right)}^2}} }} - \frac{{{x^2} - 25}}{{x - 4}} = \frac{{\left| {{{\left( {x - 5} \right)}^2}} \right|}}{{\left| {4 - x} \right|}} - \frac{{{x^2} - 25}}{{x - 4}}\\\,\,\,\,\, = \frac{{{{\left( {x - 5} \right)}^2}}}{{4 - x}} + \frac{{{x^2} - 25}}{{4 - x}} = \frac{{{{\left( {x - 5} \right)}^2} + {x^2} - 25}}{{4 - x}} = \frac{{2{x^2} - 10x}}{{4 - x}}\,\,\,\,\,(1)\end{array}$

Ta thấy $x = 3$ thỏa mãn điều kiện.

Thay $x = 3$ vào (1) ta có $C = \frac{{{{2.3}^2} - 10.3}}{{4 - 3}} = - 12$.

d) Với $x \ge 0$ ta có

$\begin{array}{l}M = 3x - \sqrt {27} + \frac{{\sqrt {{x^3} + 3{x^2}} }}{{\sqrt {x + 3} }} = 3x - \sqrt {27} + \sqrt {\frac{{{x^2}\left( {x + 3} \right)}}{{x + 3}}} = 3x - \sqrt {27} + \sqrt {{x^2}} \\\,\,\,\,\,\, = 3x - \sqrt {27} + \left| x \right| = 3x - \sqrt {27} + x = 4x - \sqrt {27} \,\,\,(1)\end{array}$

Ta thấy $x = \sqrt 3$ thỏa mãn điều kiện.

Thay $x = \sqrt 3$ vào (1) ta có $M = 4\sqrt 3 - \sqrt {27} = 4\sqrt 3 - 3\sqrt 3 = \sqrt 3$.

a) Ta có $a + \frac{1}{a} \ge 2$

$\frac{{{a^2} + 1 - 2a}}{a} \ge 0$

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

c) Với $a > 0,\,\,b > 0$ ta có

$\begin{array}{l}\sqrt a + \sqrt b \le \frac{a}{{\sqrt b }} + \frac{b}{{\sqrt a }}\\a\sqrt b + b\sqrt a \le a\sqrt a + b\sqrt b \\a\left( {\sqrt b - \sqrt a } \right) - b\left( {\sqrt b - \sqrt a } \right) \le 0\end{array}$

$$\begin{array}{l}\left( {\sqrt b - \sqrt a } \right)\left( {a - b} \right) \le 0\\\left( {\sqrt b - \sqrt a } \right)\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right) \le 0\end{array}$$

$- {\left( {\sqrt b - \sqrt a } \right)^2}\left( {\sqrt a + \sqrt b } \right) \le 0$(đúng).

b) Với $a \ge 0\;,\,\,b \ge 0$ ta có

$\sqrt {\frac{{a + b}}{2}} \ge \frac{{\sqrt a + \sqrt b }}{2}$

$\begin{array}{l}{\left( {\sqrt {\frac{{a + b}}{2}} } \right)^2} \ge {\left( {\frac{{\sqrt a + \sqrt b }}{2}} \right)^2}\\\frac{{a + b}}{2} \ge \frac{{a + b + 2\sqrt {ab} }}{4}\end{array}$

$\begin{array}{l}2\left( {a + b} \right) \ge a + b + 2\sqrt {ab} \\a + b - 2\sqrt {ab} \ge 0\end{array}$

${\left( {\sqrt a - \sqrt b } \right)^2} \ge 0$ (đúng).

d) Ta có $\frac{{{x^2} + 2}}{{\sqrt {{x^2} + 1} }} \ge 2$

$\begin{array}{l}{x^2} + 2 \ge 2\sqrt {{x^2} + 1} \\\left( {{x^2} + 1} \right) - 2\sqrt {{x^2} + 1} + 1 \ge 0\\{\left( {\sqrt {{x^2} + 1} - 1} \right)^2} \ge 0\end{array}$