Bài tập Trục căn thức ở mẫu lớp 9 (có lời giải)

Đáp án đúng là: B

$A = \frac{{\sqrt {14} - \sqrt 7 }}{{2 - \sqrt 2 }} = \frac{{\left( {\sqrt {14} - \sqrt 7 } \right)\left( {2 + \sqrt 2 } \right)}}{{\left( {2 + \sqrt 2 } \right)\left( {2 - \sqrt 2 } \right)}} = \frac{{\sqrt 7 \left( {\sqrt 2 - 1} \right)\left( {2 + \sqrt 2 } \right)}}{2}$

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

Đáp án đúng là: A

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

Đáp án đúng là: A

$B = \frac{{a\sqrt b + b\sqrt b }}{{\sqrt a + \sqrt b }} - \sqrt {ab}$

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

$B=\sqrt{ab}-\sqrt{ab}=0$

Đáp án đúng là: A

$A = \frac{{\sqrt a }}{{\sqrt a - \sqrt b }} - \frac{{\sqrt b }}{{\sqrt a + \sqrt b }}$

$A = \frac{{\sqrt a \left( {\sqrt a + \sqrt b } \right) - \sqrt b \left( {\sqrt a - \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}$

$A=\frac{a-\sqrt{ab}-\sqrt{ba}+b}{a-b}=\frac{a+b}{a-b}$

Đáp án đúng là: A

$A = \left( {\frac{{15}}{{\sqrt 6 + 1}} - \frac{4}{{\sqrt 6 - 2}} - \frac{{12}}{{3 - \sqrt 6 }}} \right)\left( {11 + \sqrt 6 } \right)$

$A = \left[ {\frac{{15\left( {\sqrt 6 - 1} \right)}}{{\left( {\sqrt 6 + 1} \right)\left( {\sqrt 6 - 1} \right)}} - \frac{{4\left( {\sqrt 6 + 2} \right)}}{{\left( {\sqrt 6 - 2} \right)\left( {\sqrt 6 + 2} \right)}} - \frac{{12\left( {3 + \sqrt 6 } \right)}}{{\left( {3 - \sqrt 6 } \right)\left( {3 + \sqrt 6 } \right)}}} \right]\left( {11 + \sqrt 6 } \right)$

$A = \left[ {\frac{{15\left( {\sqrt 6 - 1} \right)}}{5} - \frac{{4\left( {\sqrt 6 + 2} \right)}}{2} - \frac{{12\left( {3 + \sqrt 6 } \right)}}{3}} \right]\left( {11 + \sqrt 6 } \right)$

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

$A = \left[ {3\sqrt 6 - 3 - 2\sqrt 6 - 4 - 12 - 4\sqrt 6 } \right]\left( {11 + \sqrt 6 } \right)$

$A = \left( { - 3\sqrt 6 - 19} \right)\left( {11 + \sqrt 6 } \right)$

$A = - 3\sqrt 6 .11 - 3.6 - 19.11 - 19\sqrt 6$

$A=-52\sqrt{6}-227$

Đáp án đúng là: C

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

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

$A=\frac{-4}{1+\sqrt{5}}.\frac{4}{1-\sqrt{5}}=\frac{-16}{(1+\sqrt{5})(1-\sqrt{5})}=\frac{-16}{-4}=4$

Đáp án đúng là: D

$A = \left( {\frac{2}{{\sqrt 3 - 1}} + \frac{3}{{\sqrt 3 - 2}} + \frac{{15}}{{3 - \sqrt 3 }}} \right).\frac{1}{{\sqrt 3 + 5}}$

$A = \left[ {\frac{{2\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} + \frac{{3\left( {\sqrt 3 + 2} \right)}}{{\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 + 2} \right)}} + \frac{{15\left( {3 + \sqrt 3 } \right)}}{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}} \right].\frac{1}{{\sqrt 3 + 5}}$

$A = \left[ {\frac{{2\left( {\sqrt 3 + 1} \right)}}{2} + \frac{{3\left( {\sqrt 3 + 2} \right)}}{{ - 1}} + \frac{{15\left( {3 + \sqrt 3 } \right)}}{6}} \right].\frac{1}{{\sqrt 3 + 5}}$

$A = \left[ {\sqrt 3 + 1 - 3\left( {\sqrt 3 + 2} \right) + \frac{{5\left( {3 + \sqrt 3 } \right)}}{2}} \right].\frac{1}{{\sqrt 3 + 5}}$

$A = \left[ { - 2\sqrt 3 - 5 + \frac{{5\left( {3 + \sqrt 3 } \right)}}{2}} \right].\frac{1}{{\sqrt 3 + 5}}$

$A = \frac{{ - 4\sqrt 3 - 10 + 15 + 5\sqrt 3 }}{2}.\frac{1}{{\sqrt 3 + 5}}$

$A=\frac{5+\sqrt{3}}{2}.\frac{1}{\sqrt{3}+5}=\frac{1}{2}.$

Đáp án đúng là: B

$B = \left( {\frac{{\sqrt {14} - \sqrt 7 }}{{1 - \sqrt 2 }} + \frac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\frac{1}{{\sqrt 7 - \sqrt 5 }}$

$B = \left[ {\frac{{\left( {\sqrt {14} - \sqrt 7 } \right)\left( {1 - \sqrt 3 } \right) + \left( {\sqrt {15} - \sqrt 5 } \right)\left( {1 - \sqrt 2 } \right)}}{{\left( {1 - \sqrt 2 } \right)\left( {1 - \sqrt 3 } \right)}}} \right].\left( {\sqrt 7 - \sqrt 5 } \right)$

$B = \left[ {\frac{{\left( {\sqrt {14} - \sqrt 7 } \right)\left( {1 - \sqrt 3 } \right) + \sqrt 5 \left( {\sqrt 3 - 1} \right)\left( {1 - \sqrt 2 } \right)}}{{\left( {1 - \sqrt 2 } \right)\left( {1 - \sqrt 3 } \right)}}} \right].\left( {\sqrt 7 - \sqrt 5 } \right)$

$B = \left[ {\frac{{\left( {\sqrt {14} - \sqrt 7 - \sqrt 5 + \sqrt {10} } \right)\left( {1 - \sqrt 3 } \right)}}{{\left( {1 - \sqrt 2 } \right)\left( {1 - \sqrt 3 } \right)}}} \right].\left( {\sqrt 7 - \sqrt 5 } \right)$

$B = \frac{{\left( {\sqrt 7 + \sqrt 5 } \right)\left( {\sqrt 2 - 1} \right)\left( {1 - \sqrt 3 } \right)}}{{\left( {1 - \sqrt 2 } \right)\left( {1 - \sqrt 3 } \right)}}.\left( {\sqrt 7 - \sqrt 5 } \right)$

$B = - \left( {\sqrt 7 - \sqrt 5 } \right)\left( {\sqrt 7 + \sqrt 5 } \right) = - \left( {7 - 5} \right) = - 2$.

Đáp án đúng là: B

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

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

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

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

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

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

Đáp án đúng là: C

$A = \sqrt {\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }}} + \sqrt {\frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}}$

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

$A=\frac{2+\sqrt{3}}{1}+\frac{2-\sqrt{3}}{1}=2+\sqrt{3}+2-\sqrt{3}=4$

Hướng dẫn giải

Đáp án: a) Đúng. b) Đúng. c) Đúng. d) Sai.

a) Đúng.

Ta có: $m = \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} = \frac{{{m_0}}}{{\sqrt {\frac{{{c^2} - {v^2}}}{{{c^2}}}} }} = \frac{{{m_0}\sqrt {{c^2}} }}{{\sqrt {{c^2} - {v^2}} }} = \frac{{{m_0}c\sqrt {{c^2} - {v^2}} }}{{{c^2} - {v^2}}}$.

Vậy trục căn thức ở mẫu ta được $m = \frac{{{m_0}c\sqrt {{c^2} - {v^2}} }}{{{c^2} - {v^2}}}$.

b) Đúng.

Vì ${m_0} = \frac{{m\sqrt {{c^2} - {v^2}} }}{c}$ nên ${m_0}c = m\sqrt {{c^2} - {v^2}}$.

Suy ra ${m_0} = \frac{{m\sqrt {{c^2} - {v^2}} }}{c}$.

c) Đúng.

Thay $v = \frac{1}{{10}}c$ ta được:

$m = \frac{{{m_0}c\sqrt {{c^2} - {{\frac{c}{{100}}}^2}} }}{{{c^2} - \frac{{{c^2}}}{{100}}}} = \frac{{{m_0}c.\sqrt {99} c}}{{10}}:\frac{{99{c^2}}}{{100}} = \frac{{{m_0}{c^2}.\sqrt {99} }}{{10}}.\frac{{100}}{{99{c^2}}} = \frac{{10{m_0}}}{{\sqrt {99} }} \approx 1{,}005{m_0}$.

d) Sai.

Thay $v = \frac{1}{5}c$ ta được:

${m_0} = \frac{{m\sqrt {{c^2} - {v^2}} }}{c} = \frac{{m\sqrt {{c^2} - \frac{{{c^2}}}{{25}}} }}{c} = \frac{{mc\sqrt {24} }}{{5c}} = \frac{{\sqrt {24} m}}{5} \approx 0{,}98m$.

Hướng dẫn giải

Đáp án: a) Đúng. b) Đúng. c) Đúng. d) Sai.

a) Đúng.

Diện tích hình chữ nhật S$_{1}$ = $\sqrt {12} .\sqrt {18} = \sqrt {12.18} = 6\sqrt 6 .$

b) Đúng.

Diện tích hình thang ${S_2} = \frac{{\left( {\sqrt {12} + \sqrt {24} } \right).h}}{2}$.

c) Đúng.

Phương trình biểu diễn bài toán trên là: S$_{1}$ = S$_{2}$

Hay $\frac{{\left( {\sqrt {12} + \sqrt {24} } \right).h}}{2} = 6\sqrt 6$

d) Sai.

Ta có: $\frac{{\left( {\sqrt {12} + \sqrt {24} } \right).h}}{2} = 6\sqrt 6$

Suy ra $\left( {\sqrt {12} + \sqrt {24} } \right).h = 12\sqrt 6$

Do đó, $h = \frac{{12\sqrt 6 }}{{\left( {\sqrt {12} + \sqrt {24} } \right)}}$

$h = \frac{{12\sqrt 6 \left( {\sqrt {24} - \sqrt {12} } \right)}}{{\left( {\sqrt {12} + \sqrt {24} } \right)\left( {\sqrt {24} - \sqrt {12} } \right)}}$

$h = \frac{{12\sqrt 6 .\sqrt {24} - 12\sqrt 6 .\sqrt {12} }}{{{{\left( {\sqrt {24} } \right)}^2} - {{\left( {\sqrt {12} } \right)}^2}}}$

$h = \frac{{12\sqrt {6.24} - 12\sqrt {6.12} }}{{12}}$

$h = \sqrt {144} - \sqrt {72}$

$h = 12 - 6\sqrt 2$.

Hướng dẫn giải

Đáp án đúng là: a) Đúng. b) Sai. c) Sai. d) Sai.

a) Đúng.

Ta có: $a = \frac{2}{{\sqrt 3 - 1}} = \frac{{2\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}} = \frac{{2\left( {\sqrt 3 + 1} \right)}}{2} = \sqrt 3 + 1$.

b) Sai.

Ta có: $b = \frac{3}{{\sqrt 6 + \sqrt 3 }} = \frac{{3\left( {\sqrt 6 - \sqrt 3 } \right)}}{{\left( {\sqrt 6 - \sqrt 3 } \right)\left( {\sqrt 6 - \sqrt 3 } \right)}} = \frac{{3\left( {\sqrt 6 - \sqrt 3 } \right)}}{3} = \sqrt 6 - \sqrt 3$.

c) Sai.

Ta có: $c = \frac{6}{{\sqrt 6 }} = \sqrt 6$.

d) Sai.

Ta có: a + b + c = $\sqrt 3 + 1 + \sqrt 6 - \sqrt 3 + \sqrt 6 = 2\sqrt 6 + 1$.

Hướng dẫn giải

Đáp án đúng là: a) Đúng. b) Đúng. c) Đúng. d) Sai.

a) Đúng.

Ta có: $a + b = \frac{1}{{\sqrt 2 + 1}} + \frac{1}{{\sqrt 3 + \sqrt 2 }}$

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

b) Đúng.

Ta có: $c = \frac{2}{{3 + \sqrt 7 }} = \frac{{2\left( {3 - \sqrt 7 } \right)}}{{\left( {3 + \sqrt 7 } \right)\left( {3 - \sqrt 7 } \right)}} = \frac{{2\left( {3 - \sqrt 7 } \right)}}{2} = 3 - \sqrt 7$

c) Đúng.

Ta có: a + b + c = $\sqrt 3 - 1 + 3 - \sqrt 7 = 2 + \sqrt 3 - \sqrt 7$.

Vì $2 + \sqrt 3 > 2 + 1$ hay $2 + \sqrt 3 > 3$ nên $2 + \sqrt 3 - \sqrt 7 > 3 - \sqrt 7 = \sqrt 9 - \sqrt 7 > 0$

Do đó, a + b + c > 0.

d) Sai.

Ta có: $T = {\left( {a + b + c + 4 - \sqrt 7 } \right)^2} = {\left( {2 + \sqrt 3 - \sqrt 7 + 4 - \sqrt 7 } \right)^2} = {\left( {6 + \sqrt 3 } \right)^2} = 39 + 12\sqrt 3$.

Do đó T không có giá trị là số nguyên.

Hướng dẫn giải

Đáp án đúng là: a) Đúng. b) Đúng. c) Sai. d) Sai.

a) Đúng.

Ta có: $a = \frac{4}{{3 - \sqrt 5 }} = \frac{{4\left( {3 + \sqrt 5 } \right)}}{{\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)}} = \frac{{4\left( {3 + \sqrt 5 } \right)}}{4} = 3 + \sqrt 5$.

b) Đúng.

Ta có: $b = \frac{{\sqrt {10} + \sqrt 2 }}{{\sqrt 5 + 1}} = \frac{{\sqrt 2 \left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 + 1}} = \sqrt 2$.

c) Sai.

Ta có: $c = \frac{{2\sqrt 2 }}{{\sqrt 6 + 2}} = \frac{{2\sqrt 2 }}{{\sqrt 2 \left( {\sqrt 3 + \sqrt 2 } \right)}} = \frac{2}{{\sqrt 3 + \sqrt 2 }}$

Vì $\sqrt 2 + \sqrt 3 > 2$ nên $\frac{2}{{\sqrt 3 + \sqrt 2 }} < 1 < \sqrt 2$ nên c < b.

d) Sai.

Ta có:

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

Do đó, a + b + c có giá trị là số vô tỉ.

Hướng dẫn giải

Đáp án: 8

$B = \frac{3}{{\sqrt 5 - \sqrt 2 }} + \frac{4}{{\sqrt 6 + \sqrt 2 }} + \frac{1}{{\sqrt 6 + \sqrt 5 }}$

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

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

$= \frac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{3} + \frac{{4\left( {\sqrt 6 - \sqrt 2 } \right)}}{4} + \frac{{\sqrt 6 - \sqrt 5 }}{1}$

$= \sqrt 5 + \sqrt 2 + \sqrt 6 - \sqrt 2 + \sqrt 6 - \sqrt 5$

$= 2\sqrt 6$.

Do đó, a = 2, b = 6.

Vậy T = a + b = 8.

Hướng dẫn giải

Đáp án: 4

$D = \left( {\frac{{\sqrt {22} - \sqrt {11} }}{{1 - \sqrt 2 }} + \frac{{\sqrt {21} - \sqrt 7 }}{{1 - \sqrt 3 }}} \right)\left( {\sqrt 7 - \sqrt {11} } \right)$

$$= \left[ {\frac{{\left( {\sqrt {22} - \sqrt {11} } \right)\left( {1 + \sqrt 2 } \right)}}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}} + \frac{{\left( {\sqrt {21} - \sqrt 7 } \right)\left( {\sqrt 3 + 1} \right)}}{{\left( {1 - \sqrt 3 } \right)\left( {\sqrt 3 + 1} \right)}}} \right]\left( {\sqrt 7 - \sqrt {11} } \right)$$

$$= \left[ {\frac{{\left( {\sqrt {22} - \sqrt {11} } \right)\left( {1 + \sqrt 2 } \right)}}{{ - 1}} - \frac{{\left( {\sqrt {21} - \sqrt 7 } \right)\left( {\sqrt 3 + 1} \right)}}{2}} \right]\left( {\sqrt 7 - \sqrt {11} } \right)$$

$$= \left[ { - \left( {\sqrt {44} - \sqrt {22} + \sqrt {22} - \sqrt {11} } \right) - \frac{{\sqrt {21} - \sqrt 7 + \sqrt {63} - \sqrt {21} }}{2}} \right]\left( {\sqrt 7 - \sqrt {11} } \right)$$

$$= \left[ { - \sqrt {44} + \sqrt {11} - \frac{{\sqrt 7 \left( {\sqrt 9 - 1} \right)}}{2}} \right]\left( {\sqrt 7 - \sqrt {11} } \right)$$

$$= \left[ { - 2\sqrt {11} + \sqrt {11} - \sqrt 7 } \right]\left( {\sqrt 7 - \sqrt {11} } \right)$$

$$= \left( { - \sqrt {11} - \sqrt 7 } \right)\left( {\sqrt 7 - \sqrt {11} } \right)$$

$$= - \left( {\sqrt {11} + \sqrt 7 } \right)\left( {\sqrt 7 - \sqrt {11} } \right)$$

$$= - \left[ {{{\left( {\sqrt 7 } \right)}^2} - {{\left( {\sqrt {11} } \right)}^2}} \right]$$

$= - \left( { - 4} \right) = 4$.

Hướng dẫn giải

Đáp án: −0,4

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

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

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

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

$= \frac{{4\sqrt {xy} - x - 2\sqrt {xy} - y}}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}.\frac{{2\sqrt x }}{{\sqrt x - \sqrt y }}$

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

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

$= \frac{{ - \sqrt x }}{{\sqrt x + \sqrt y }}$

Với $\frac{x}{y} = \frac{4}{9}$ suy ra y = $\frac{{9x}}{4}$.

Do đó, P $= \frac{{ - \sqrt x }}{{\sqrt x + \sqrt y }} = \frac{{ - \sqrt x }}{{\sqrt x + \frac{3}{2}\sqrt x }} = \frac{{ - \sqrt x }}{{\frac{5}{2}\sqrt x }} = - \frac{2}{5} = - 0{,}4$.

Hướng dẫn giải

Đáp án: 2

Ta có: $\frac{{13}}{{3\sqrt 7 + \sqrt {11} }} + \frac{{17}}{{4\sqrt 7 + 2\sqrt {11} }}$

$= \frac{{13\left( {3\sqrt 7 - \sqrt {11} } \right)}}{{\left( {3\sqrt 7 + \sqrt {11} } \right)\left( {3\sqrt 7 - \sqrt {11} } \right)}} + \frac{{17\left( {4\sqrt 7 + 2\sqrt {11} } \right)}}{{\left( {4\sqrt 7 + 2\sqrt {11} } \right)\left( {4\sqrt 7 - 2\sqrt {11} } \right)}}$

$= \frac{{13\left( {3\sqrt 7 - \sqrt {11} } \right)}}{{52}} + \frac{{17\left( {4\sqrt 7 + 2\sqrt {11} } \right)}}{{68}}$

$= \frac{{3\sqrt 7 - \sqrt {11} }}{4} + \frac{{4\sqrt 7 + 2\sqrt {11} }}{4}$

$= \frac{{3\sqrt 7 - \sqrt {11} + 4\sqrt 7 + 2\sqrt {11} }}{4}$

$= \frac{{7\sqrt 7 + \sqrt {11} }}{4}$

$= \frac{{7\sqrt 7 }}{4} + \frac{{\sqrt {11} }}{4} = \frac{7}{4}.\sqrt 7 + \frac{1}{4}.\sqrt {11}$.

Do đó, $a = \frac{7}{4};\,\,b = \frac{1}{4}$.

Suy ra a + b = $\frac{7}{4} + \frac{1}{4} = 2.$

Hướng dẫn giải

Đáp án: 0

Ta có: $x = \frac{{3 + \sqrt 2 }}{{2\sqrt 2 - 1}} = \frac{{\left( {3 + \sqrt 2 } \right)\left( {2\sqrt 2 + 1} \right)}}{{\left( {2\sqrt 2 - 1} \right)\left( {2\sqrt 2 + 1} \right)}} = \frac{{6\sqrt 2 + 3 + 4 + \sqrt 2 }}{7} = \frac{{7\sqrt 2 + 7}}{7} = \sqrt 2 + 1$.

Ta có: M = x$^{5}$ – 6x$^{3}$ + x

$= {\left( {\sqrt 2 + 1} \right)^5} - 6.{\left( {\sqrt 2 + 1} \right)^3} + \left( {\sqrt 2 + 1} \right)$

$= \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {\sqrt 2 + 1} \right)}^4} - 6{{\left( {\sqrt 2 + 1} \right)}^2} + 1} \right]$

$= \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {\sqrt 2 + 1} \right)}^4} - 2{{\left( {\sqrt 2 + 1} \right)}^2} + 1 - 4{{\left( {\sqrt 2 + 1} \right)}^2}} \right]$

$= \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {\sqrt 2 + 1} \right)}^4} - 2{{\left( {\sqrt 2 + 1} \right)}^2} + 1 - 4{{\left( {\sqrt 2 + 1} \right)}^2}} \right]$

$= \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {{{\left( {\sqrt 2 + 1} \right)}^2} - 1} \right)}^2} - 4{{\left( {\sqrt 2 + 1} \right)}^2}} \right]$

$= \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {\sqrt 2 + 1} \right)}^2} - 2\left( {\sqrt 2 + 1} \right) - 1} \right]\left[ {{{\left( {\sqrt 2 + 1} \right)}^2} + 2\left( {\sqrt 2 + 1} \right) - 1} \right]$

$= \left( {\sqrt 2 + 1} \right)\left[ {2 + 2\sqrt 2 + 1 - 2\sqrt 2 - 2 - 1} \right]\left[ {{{\left( {\sqrt 2 + 1} \right)}^2} + 2\left( {\sqrt 2 + 1} \right) - 1} \right]$

$= 0.\left( {\sqrt 2 + 1} \right).\left[ {{{\left( {\sqrt 2 + 1} \right)}^2} + 2\left( {\sqrt 2 + 1} \right) - 1} \right]$

= 0.