A11= ( 1/căn a -1 - 1 / căn a ) : ( căn a +1 / căn a -2 - văn a +2/căn a -1). 1 Tìm điều kiện và rút gọn 2 Tính A11 khi x=7-4căn3 3 Tìm x để A11 = 1/4 4 Tìm x để A11>0
1 câu trả lời
Đáp án:
$\begin{array}{l}
1)Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
a \ne 0\\
\sqrt a \ne 1\\
\sqrt a \ne 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
a \ne 1\\
a \ne 4
\end{array} \right.\\
Vậy\,a > 0;a \ne 1;a \ne 4\\
2)\\
A = \left( {\dfrac{1}{{\sqrt a - 1}} - \dfrac{1}{{\sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 1}}{{\sqrt a - 2}} - \dfrac{{\sqrt a + 2}}{{\sqrt a - 1}}} \right)\\
= \dfrac{{\sqrt a - \left( {\sqrt a - 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}}:\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}{{a - 1 - a + 4}}\\
= \dfrac{1}{{\sqrt a }}.\dfrac{{\sqrt a - 2}}{3}\\
= \dfrac{{\sqrt a - 2}}{{3\sqrt a }}\\
2)x = 7 - 4\sqrt 3 \left( {tmdk} \right)\\
= 4 - 2.2\sqrt 3 + 3\\
= {\left( {2 - \sqrt 3 } \right)^2}\\
\Leftrightarrow \sqrt x = 2 - \sqrt 3 \\
\Leftrightarrow A = \dfrac{{\sqrt a - 2}}{{3\sqrt a }} = \dfrac{{2 - \sqrt 3 - 2}}{{3\left( {2 - \sqrt 3 } \right)}}\\
= \dfrac{{ - \sqrt 3 }}{{3\left( {2 - \sqrt 3 } \right)}}\\
= \dfrac{1}{{\sqrt 3 \left( {\sqrt 3 - 2} \right)}}\\
= \dfrac{1}{{3 - 2\sqrt 3 }}\\
= \dfrac{{3 + 2\sqrt 3 }}{{{3^2} - {{\left( {2\sqrt 3 } \right)}^2}}}\\
= \dfrac{{ - 3 - 2\sqrt 3 }}{3}\\
3)A = \dfrac{1}{4}\\
\Leftrightarrow \dfrac{{\sqrt a - 2}}{{3\sqrt a }} = \dfrac{1}{4}\\
\Leftrightarrow 4\sqrt a - 8 = 3\sqrt a \\
\Leftrightarrow \sqrt a = 8\\
\Leftrightarrow a = 64\left( {tmdk} \right)\\
Vậy\,a = 64\\
4)A > 0\\
\Leftrightarrow \dfrac{{\sqrt a - 2}}{{3\sqrt a }} > 0\\
\Leftrightarrow \sqrt a - 2 > 0\\
\Leftrightarrow \sqrt a > 2\\
\Leftrightarrow a > 4\\
Vậy\,a > 4
\end{array}$