Toán Lớp 9: 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
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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}$