Toán Lớp 9: Cho biểu thức Q = ($\frac{\sqrt{1}}{\sqrt{x-2}}$ + $\frac{5\sqrt{x}-4 }{2\sqrt{x}-x }$ ) : ($\frac{2+\sqrt{x}}{\sqrt{x}}$ – $\frac{\sqrt{x}}{x-2}$
a) Rút gọn Q
b) Tính giá trị của Q tại x = $\frac{3-\sqrt{5}}{2}$
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Toán Lớp 9: Cho biểu thức Q = ($\frac{\sqrt{1}}{\sqrt{x-2}}$ + $\frac{5\sqrt{x}-4 }{2\sqrt{x}-x }$ ) : ($\frac{2+\sqrt{x}}{\sqrt{x}}$ – $\frac{\sqrt{x}}{x-2}$
a) Rút gọn Q
b) Tính giá trị của Q tại x = $\frac{3-\sqrt{5}}{2}$
Comments ( 1 )
a)Dkxd:x > 0;x \ne 4\\
Q = \left( {\dfrac{1}{{\sqrt x – 2}} + \dfrac{{5\sqrt x – 4}}{{2\sqrt x – x}}} \right):\left( {\dfrac{{2 + \sqrt x }}{{\sqrt x }} – \dfrac{{\sqrt x }}{{\sqrt x – 2}}} \right)\\
= \dfrac{{\sqrt x – 5\sqrt x + 4}}{{\sqrt x \left( {\sqrt x – 2} \right)}}:\dfrac{{\left( {2 + \sqrt x } \right)\left( {\sqrt x – 2} \right) – \sqrt x .\sqrt x }}{{\sqrt x \left( {\sqrt x – 2} \right)}}\\
= \dfrac{{ – 4 – 4\sqrt x }}{{\sqrt x \left( {\sqrt x – 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x – 2} \right)}}{{x – 4 – x}}\\
= \dfrac{{ – 4\left( {\sqrt x + 1} \right)}}{{ – 4}}\\
= \sqrt x + 1\\
b)x = \dfrac{{3 – \sqrt 5 }}{2}\left( {tmdk} \right)\\
= \dfrac{{6 – 2\sqrt 5 }}{4}\\
= {\left( {\dfrac{{\sqrt 5 – 1}}{2}} \right)^2}\\
\Leftrightarrow \sqrt x = \dfrac{{\sqrt 5 – 1}}{2}\\
\Leftrightarrow Q = \sqrt x + 1 = \dfrac{{\sqrt 5 – 1}}{2} + 1 = \dfrac{{\sqrt 5 + 1}}{2}
\end{array}$