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

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

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

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

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

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

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

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

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

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

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