Toán Lớp 9: Cho A = ( √x + 2 $\frac{ √x + 2}{x – 1}$ – $\frac{√x – 2}{x – 2 √x + 1}$ ) : $\frac{4x}{ (x – 1) ²}$
a, Rút gọn P
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Toán Lớp 9: Cho A = ( √x + 2 $\frac{ √x + 2}{x – 1}$ – $\frac{√x – 2}{x – 2 √x + 1}$ ) : $\frac{4x}{ (x – 1) ²}$
a, Rút gọn P
Comments ( 1 )
Dk{\rm{xd:x > 0;x}} \ne {\rm{1}}\\
A = \left( {\dfrac{{\sqrt x + 2}}{{x – 1}} – \dfrac{{\sqrt x – 2}}{{x – 2\sqrt x + 1}}} \right):\dfrac{{4x}}{{{{\left( {x – 1} \right)}^2}}}\\
= \left( {\dfrac{{\sqrt x + 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}} – \dfrac{{\sqrt x – 2}}{{{{\left( {\sqrt x – 1} \right)}^2}}}} \right).\dfrac{{{{\left( {x – 1} \right)}^2}}}{{4x}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right) – \left( {\sqrt x – 2} \right)\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x – 1} \right)}^2}\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {\sqrt x – 1} \right)}^2}{{\left( {\sqrt x + 1} \right)}^2}}}{{4x}}\\
= \dfrac{{x + \sqrt x – 2 – \left( {x – \sqrt x – 2} \right)}}{1}.\dfrac{{\sqrt x + 1}}{{4x}}\\
= \dfrac{{2\sqrt x }}{1}.\dfrac{{\sqrt x + 1}}{{4x}}\\
= \dfrac{{\sqrt x + 1}}{{2\sqrt x }}
\end{array}$