Đáp án:+Giải thích các bước giải:
a) ĐKXĐ: $\begin{cases} x \ge 0\\ \sqrt{x}-1 \ne 0\\ \sqrt{x}+1 \ne 0\\ x-1 \ne 0\end{cases}$ `=>x>=0; x \ne 1`
`A=\sqrtx/(\sqrtx-1)+3/(\sqrtx+1)-(6\sqrtx-4)/(x-1)` `=(\sqrtx(\sqrtx+1))/((\sqrtx+1)(\sqrtx-1))+(3(\sqrtx-1))/((\sqrtx+1)(\sqrtx-1))-(6\sqrtx-4)/((\sqrtx+1)(\sqrtx-1))` `=(\sqrtx(\sqrtx+1)+3(\sqrtx-1)-(6\sqrtx-4))/((\sqrtx+1)(\sqrtx-1))` `=(x+\sqrtx+3\sqrtx-3-6\sqrtx+4)/((\sqrtx+1)(\sqrtx-1))` `=(x-2\sqrtx+1)/((\sqrtx+1)(\sqrtx-1))` `=(\sqrtx-1)^2/((\sqrtx+1)(\sqrtx-1))=(\sqrtx-1)/(\sqrtx+1)`
b) `x=3-2\sqrt2=(\sqrt2-1)^2` `=> \sqrtx=\sqrt((\sqrt2-1)^2)=|\sqrt2-1|=\sqrt2-1`
`=>A=(\sqrtx-1)/(\sqrtx+1)=(\sqrt2-1-1)/(\sqrt2-1+1)=(\sqrt2-2)/\sqrt2=(\sqrt2(1-\sqrt2))/\sqrt2=1-\sqrt2`
c) `A<1/2 <=> (\sqrtx-1)/(\sqrtx+1)<1/2`
`<=> (\sqrtx-1)/(\sqrtx+1)-1/2<0`
`<=> (2(\sqrtx-1)-(\sqrtx+1))/(2(\sqrtx+1))<0`
`<=> (2\sqrtx-2-\sqrtx-1)/(2(\sqrtx+1))<0`
`<=> (\sqrtx-3)/(2(\sqrtx+1))<0`
`=> \sqrtx-3 < 0` (do `2(\sqrtx+1) > 0` `∀ x >= 0`)
`<=> \sqrtx < 3 <=> x < 9`
Kết hợp đk `x ≥ 0; x \ne 1`
`=> 0 ≤ x < 9; x \ne 1`
