B=(1/căn+1-1/x+căn x):x-căn x+1/x căn x+1 (x>0)
C=(x+1/căn x +2)nhân căn x/(căn x+1)(x căn x +1)
D=(x căn x-1/căn x-1 +căn x):(x-1)-2/căn x-1 (với x>bằng 0,x khác 1)
B=(1/căn+1-1/x+căn x):x-căn x+1/x căn x+1 (x>0)
C=(x+1/căn x +2)nhân căn x/(căn x+1)(x căn x +1)
D=(x căn x-1/căn x-1 +căn x):(x-1)-2/căn x-1 (với x>bằng 0,x khác 1)
Đáp án:
$\begin{array}{l}
B = \left( {\dfrac{1}{{\sqrt x + 1}} – \dfrac{1}{{x + \sqrt x }}} \right):\dfrac{{x – \sqrt x + 1}}{{x\sqrt x + 1}}\\
= \dfrac{{\sqrt x – 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}:\dfrac{{x – \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x – 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{\sqrt x – 1}}{{\sqrt x }}\\
C = \left( {\dfrac{{x + 1}}{{\sqrt x }} + 2} \right).\dfrac{{\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {x – \sqrt x + 1} \right)}}\\
= \dfrac{1}{{x – \sqrt x + 1}}\\
D = \left( {\dfrac{{x\sqrt x – 1}}{{\sqrt x – 1}} + \sqrt x } \right):\left( {x – 1} \right) – \dfrac{2}{{\sqrt x – 1}}\\
= \left( {\dfrac{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x – 1}} + \sqrt x } \right).\dfrac{1}{{x – 1}} – \dfrac{2}{{\sqrt x – 1}}\\
= \left( {x + 2\sqrt x + 1} \right).\dfrac{1}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}} – \dfrac{2}{{\sqrt x – 1}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} – \dfrac{2}{{\sqrt x – 1}}\\
= \dfrac{{\sqrt x – 1}}{{\sqrt x – 1}}\\
= 1
\end{array}$