Rút gọn $\frac{1}{2\sqrt[]{x}-2}$ + $\frac{1}{2\sqrt[]{x}+2}$ + $\frac{x}{1-x}$ với x $\geq$ 0 , x$\neq$ 1 16/07/2021 Bởi Lyla Rút gọn $\frac{1}{2\sqrt[]{x}-2}$ + $\frac{1}{2\sqrt[]{x}+2}$ + $\frac{x}{1-x}$ với x $\geq$ 0 , x$\neq$ 1
`1/(2sqrt(x)-2) + 1/(2sqrt(x)+2) + x/(1-x) \ \ (ĐK : x>=0 ; x ne 1)` `= 1/(2(sqrt(x)-1)) + 1/(2(sqrt(x)+1)) – x/(x+1)` `= (sqrt(x)+1+sqrt(x)-1-2x)/(2(sqrt(x)-1)(sqrt(x)+1))` `= (2sqrt(x)-2x)/(2(x-1))=(2(sqrt(x)-x))/(2(x-1))` `= (sqrt(x)-x)/(x-1)` Bình luận
`1/(2sqrt(x)-2) + 1/(2sqrt(x)+2) + x/(1-x) \ \ (ĐK : x>=0 ; x ne 1)`
`= 1/(2(sqrt(x)-1)) + 1/(2(sqrt(x)+1)) – x/(x+1)`
`= (sqrt(x)+1+sqrt(x)-1-2x)/(2(sqrt(x)-1)(sqrt(x)+1))`
`= (2sqrt(x)-2x)/(2(x-1))=(2(sqrt(x)-x))/(2(x-1))`
`= (sqrt(x)-x)/(x-1)`
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