Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
P\(=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)
\(=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-1}{\sqrt{x}+1}\)
a)\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\\ A=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ A=\frac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)-2\sqrt{x}}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ A=\frac{1}{\sqrt{x}+1}\)