Rút gọn: B= $\frac{x+\sqrt{x}}{1-x}$ + $\frac{(\sqrt{x-2 })^2-\sqrt{x}-x }{1-\sqrt{x}}$

2 câu trả lời

$\displaystyle \begin{array}{{>{\displaystyle}l}} B=\frac{x+\sqrt{x}}{1-x} +\frac{\left(\sqrt{x-2}\right)^{2} -\sqrt{x} -x}{1-\sqrt{x}}\\ DKXD\ :\ x\geqslant 0;x\#1\\ =\frac{\sqrt{x}\left(\sqrt{x} +1\right)}{\left( 1-\sqrt{x}\right)\left( 1+\sqrt{x}\right)} +\frac{x-2-\sqrt{x} -x}{1-\sqrt{x}}\\ =\frac{\sqrt{x} -2-\sqrt{x}}{1-\sqrt{x}} =\frac{-2}{1-\sqrt{x}} \end{array}$

`#huy`

`ĐKXĐ:{(x>=0),(x\ne1):}`

`B=(x+\sqrt{x})/(1-x)+(\sqrt{(x-2)^2}-\sqrt{x}-x)/(1-\sqrt{x})`

`=(\sqrt{x}(\sqrt{x}+1))/((1-\sqrt{x})(1+\sqrt{x}))+(x-2-\sqrt{x}-x)/(1-\sqrt{x})`

`=(\sqrt{x}-2-\sqrt{x})/(1-\sqrt{x})`

`=(-2)/(1-\sqrt{x})`

 

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