Rút gọn A chỗ 1 phần căn x - 1 - 2 phần x-1 là chia nha mn

Đáp án:

$A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$

Giải thích các bước giải:

$A=\dfrac{\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}}{\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}}\,\,\,(x\ge0;x\ne1)\\=\dfrac{\dfrac{1}{\sqrt{x}+1}-\dfrac{2(\sqrt{x}-1)}{\sqrt{x}(x-1)+(x-1)}}{\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}}\\=\dfrac{\dfrac{1}{\sqrt{x}+1}-\dfrac{2(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}}{\dfrac{\sqrt{x}+1-2}{(\sqrt{x}-1)(\sqrt{x}+1)}}\\=\dfrac{\dfrac{\sqrt{x}+1-2}{(\sqrt{x}+1)^2}}{\dfrac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)}}\\=\dfrac{\sqrt{x}-1}{(\sqrt{x}+1)^2}:\dfrac{1}{\sqrt{x}+1}\\=\dfrac{\sqrt{x}-1}{(\sqrt{x}+1)^2}.(\sqrt{x}+1)\\=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$