\(\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\left(\frac{x-1}{\sqrt{x}+1}-2\right)\left(ĐK:x\ne\pm1\right)\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x-1-2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x-2\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\left(\sqrt{x}-3\right)}{x-1}\)
\(=\frac{2\sqrt{x}-6}{x-1}\)
\(\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\left(\frac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1}{\sqrt{x}+1}-\frac{2\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1-2\sqrt{x}-2}{\sqrt{x}+1}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-2\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(=\left(\frac{2\sqrt{x}\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\)
\(=\frac{2x\sqrt{x}-6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)
Bài lm có sai sót chỗ nào thì mong bn sửa lại........
Mk viết thiếu \(\sqrt{x}\)từ bước thứ 4 nhé bạn sửa r` lm tiếp là ra
\(\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\left(\frac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right)\left(\frac{x-1-2\sqrt{x}-2}{\sqrt{x}+1}\right)\)
\(=\left(\frac{2\sqrt{x}}{x-1}\right)\left(\frac{x-2\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(=\frac{2x\sqrt{x}-2x-6\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
