Toán (x+2 /x+căn x-2 – căn x /căn x +2) : 1 / x-1 mọng mọi người giúp đỡ 20/09/2021 By Hailey (x+2 /x+căn x-2 – căn x /căn x +2) : 1 / x-1 mọng mọi người giúp đỡ
Đáp án: $\begin{array}{l}Dkxd:x \ge 0;x \ne 1\\\left( {\dfrac{{x + 2}}{{x + \sqrt x – 2}} – \dfrac{{\sqrt x }}{{\sqrt x + 2}}} \right):\dfrac{1}{{x – 1}}\\ = \left[ {\dfrac{{x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}} – \dfrac{{\sqrt x }}{{\sqrt x + 2}}} \right].\left( {x – 1} \right)\\ = \dfrac{{x + 2 – \sqrt x \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}.\left( {\sqrt x – 1} \right).\left( {\sqrt x + 1} \right)\\ = \dfrac{{x + 2 – x + \sqrt x }}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\ = \sqrt x + 1\end{array}$ Trả lời
Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
\left( {\dfrac{{x + 2}}{{x + \sqrt x – 2}} – \dfrac{{\sqrt x }}{{\sqrt x + 2}}} \right):\dfrac{1}{{x – 1}}\\
= \left[ {\dfrac{{x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}} – \dfrac{{\sqrt x }}{{\sqrt x + 2}}} \right].\left( {x – 1} \right)\\
= \dfrac{{x + 2 – \sqrt x \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}.\left( {\sqrt x – 1} \right).\left( {\sqrt x + 1} \right)\\
= \dfrac{{x + 2 – x + \sqrt x }}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\
= \sqrt x + 1
\end{array}$