Toán Lớp 9: A= (2/x-√x – 1/√x-1) : x-4/x√x+√x – 2x với x>0, x khác 1, x khác 4
a) rút gọn A
b) tìm x để A > -1/2
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Toán Lớp 9: A= (2/x-√x – 1/√x-1) : x-4/x√x+√x – 2x với x>0, x khác 1, x khác 4
a) rút gọn A
b) tìm x để A > -1/2
Comments ( 1 )
Dkxd:x > 0;x \ne 1;x \ne 4\\
a)A = \left( {\dfrac{2}{{x – \sqrt x }} – \dfrac{1}{{\sqrt x – 1}}} \right):\dfrac{{x – 4}}{{x\sqrt x + \sqrt x – 2x}}\\
= \dfrac{{2 – \sqrt x }}{{\sqrt x \left( {\sqrt x – 1} \right)}}.\dfrac{{\sqrt x \left( {x – 2\sqrt x + 1} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{ – 1}}{{\sqrt x – 1}}.\dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\sqrt x + 2}}\\
= \dfrac{{1 – \sqrt x }}{{\sqrt x + 2}}\\
b)A > – \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{1 – \sqrt x }}{{\sqrt x + 2}} + \dfrac{1}{2} > 0\\
\Leftrightarrow \dfrac{{2 – 2\sqrt x + \sqrt x + 2}}{{2\left( {\sqrt x + 2} \right)}} > 0\\
\Leftrightarrow 4 – \sqrt x > 0\\
\Leftrightarrow \sqrt x < 4\\
\Leftrightarrow x < 16\\
Vậy\,0 < x < 16;x \ne 1;x \ne 4
\end{array}$