Rút gọn biểu thức : $A$ = ( $\frac{12}{\sqrt{x}-1}$ +$\frac{12}{1+\sqrt{x}}$ ):$\frac{2017}{x-1}$ với ($x$ $\geq$ 0; x$\neq$ 1)

`A=(12/(\sqrt{x}-1)+12/(1+\sqrt{x})):2017/(x-1)`

`ĐKXĐ:x>0n; x\ne 1`

`<=>A=(12/(\sqrt{x}-1)+12/(\sqrt{x}+1)):2017/((\sqrt{x})²-1²)`

`<=>A=(12/(\sqrt{x}-1)+12/(\sqrt{x}+1)):2017/((\sqrt{x}-1)(\sqrt{x}+1))`

`<=>A=((12(\sqrt{x}+1))/((\sqrt{x}-1)(\sqrt{x}+1))+(12(\sqrt{x}-1))/((\sqrt{x}-1)(\sqrt{x}+1)) ):2017/((\sqrt{x}-1)(\sqrt{x}+1))`

`<=>A=((12\sqrt{x}+12)/((\sqrt{x}-1)(\sqrt{x}+1))+(12\sqrt{x}-12)/((\sqrt{x}-1)(\sqrt{x}+1))):2017/((\sqrt{x}-1)(\sqrt{x}+1))`

`<=>A=(12\sqrt{x}+12+12\sqrt{x}-12)/((\sqrt{x}-1)(\sqrt{x}+1)):2017/((\sqrt{x}-1)(\sqrt{x}+1))`

`<=>A=((12+12)\sqrt{x}+(12-12))/((\sqrt{x}-1)(\sqrt{x}+1)). ((\sqrt{x}-1)(\sqrt{x}+1))/2017`

`<=>A=(24\sqrt{x})/((\sqrt{x}-1)(\sqrt{x}+1)).((\sqrt{x}-1)(\sqrt{x}+1))/2017`

`<=>A=(24\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1))/((\sqrt{x}-1)(\sqrt{x}+1).2017)`

`<=>A=(24\sqrt{x})/2017` ( giản ước `(\sqrt{x}-1)(\sqrt{x}+1)` )

Vậy `A=(24\sqrt{x})/2017`