Toán Lớp 9: rút gọn P = (x+3/x-1 + √x/1- √x + 1/ 1+ √x).(x √x-1/√x-1 + √x) x ≥0,x khác 1
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Toán Lớp 9: rút gọn P = (x+3/x-1 + √x/1- √x + 1/ 1+ √x).(x √x-1/√x-1 + √x) x ≥0,x khác 1
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P = \left( {\dfrac{{x + 3}}{{x – 1}} + \dfrac{{\sqrt x }}{{1 – \sqrt x }} + \dfrac{1}{{1 + \sqrt x }}} \right).\left( {\dfrac{{x\sqrt x – 1}}{{\sqrt x – 1}} + \sqrt x } \right)\\
= \dfrac{{x + 3 – \sqrt x \left( {1 + \sqrt x } \right) + \sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}.\left[ {\dfrac{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x – 1}} + \sqrt x } \right]\\
= \dfrac{{x + 3 – \sqrt x – x + \sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}.\left( {x + \sqrt x + 1 + \sqrt x } \right)\\
= \dfrac{2}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}.{\left( {\sqrt x + 1} \right)^2}\\
= \dfrac{{2\left( {\sqrt x + 1} \right)}}{{\sqrt x – 1}}
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