A=(((√X -2)/(X-1))-((√X +2)/(X+2√X +1))).(X ²-2X+1)/2
a)Rút gọn A
GIÚP MK VS . MK ĐANG CẦN GẤP . CÒN 15′ NỮA LÀ NỘP BÀI R Ạ GIÚP EM VS
A=(((√X -2)/(X-1))-((√X +2)/(X+2√X +1))).(X ²-2X+1)/2
a)Rút gọn A
GIÚP MK VS . MK ĐANG CẦN GẤP . CÒN 15′ NỮA LÀ NỘP BÀI R Ạ GIÚP EM VS
Đáp án:
$Dkxd:x \ge 0;x \ne 1$
$\begin{array}{l}
A = \left( {\frac{{\sqrt x – 2}}{{x – 1}} – \frac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\frac{{{x^2} – 2x + 1}}{2}\\
= \frac{{\sqrt x – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}} – \frac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}.\frac{{{{\left( {x – 1} \right)}^2}}}{2}\\
= \frac{{\left( {\sqrt x – 2} \right).\left( {\sqrt x + 1} \right) – \left( {\sqrt x + 2} \right).\left( {\sqrt x – 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}.\left( {\sqrt x – 1} \right)}}.\frac{{{{\left( {\sqrt x – 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{2}\\
= \frac{{x – \sqrt x – 2 – x – \sqrt x + 2}}{1}.\frac{{\sqrt x – 1}}{2}\\
= \frac{{ – 2\sqrt x }}{1}.\frac{{\sqrt x – 1}}{2}\\
= – \sqrt x \left( {\sqrt x – 1} \right)\\
= \sqrt x – x
\end{array}$