Toán Lớp 9: P = ( x – √ x + 2 / x- 1 – 1 / √ x – 1 ) . X + 2 √ x + 1 / 2x – 2 √ x . Giúp em với a em đang cần gấp
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Toán Lớp 9: P = ( x – √ x + 2 / x- 1 – 1 / √ x – 1 ) . X + 2 √ x + 1 / 2x – 2 √ x . Giúp em với a em đang cần gấp
Comments ( 1 )
DK:x > 0;x \ne 1\\
P = \left( {\dfrac{{x – \sqrt x + 2}}{{x – 1}} – \dfrac{1}{{\sqrt x – 1}}} \right).\dfrac{{x + 2\sqrt x + 1}}{{2x – 2\sqrt x }}\\
= \dfrac{{x – \sqrt x + 2 – \sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{2\sqrt x \left( {\sqrt x – 1} \right)}}\\
= \dfrac{{x – 2\sqrt x + 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{2\sqrt x \left( {\sqrt x – 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{2\sqrt x \left( {\sqrt x – 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{2\sqrt x }}
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