Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{x-1}\)
\(=2+\dfrac{2x+2}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}-x-3-\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{-x-4}=\frac{-4\sqrt{x}}{-x-4}=\frac{4\sqrt{x}}{x+4}\)
dk: x>=0 và x khác 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
ĐKXĐ:
\(x\ge0\text{ và }\sqrt{x}-1\ge0\text{ và }x+1\ne0\)
\(\Leftrightarrow x\ge1\text{ và }x\ne-1\)
\(\left(\frac{x\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{x+1}\right):\frac{\sqrt{x}-1}{x+1}\)
\(=\left(\frac{x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}+1}\)
\(=\left(\frac{x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right).\frac{x+1}{\sqrt{x}+1}\)
\(=\frac{x\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}.\frac{x+1}{\sqrt{x}+1}\)
\(=\frac{x\sqrt{x}+\sqrt{x}+1}{x+2\sqrt{x}+1}\)
ko bik làm hay làm biến đó