giúp mk với mk cần gấp

\(\sqrt{x}=y\\ \)

ĐK: \(x\ne0,1,4\Leftrightarrow\left\{\begin{matrix}y>0\\y\ne1\&4\end{matrix}\right.\) ko sửa được y khác 1 &2

\(P=\left(\frac{\left(1-y\right)}{\left(y-2\right)}+\frac{y}{\left(y-1\right)}+\frac{y+2}{\left(y-1\right)\left(y-2\right)}\right):\left(\frac{2}{y-2}-\frac{y-1}{y\left(y-2\right)}\right)\)

\(P=\left(\frac{2y-y^2-1}{\left(y-2\right)\left(y-1\right)}+\frac{y^2-2y}{\left(y-1\right)\left(y-2\right)}+\frac{y+2}{\left(y-1\right)\left(y-2\right)}\right):\left(\frac{2y-y+1}{y\left(y-2\right)}\right)\)

\(P=\left(\frac{y+1}{\left(y-1\right)\left(y-2\right)}\right).\left(\frac{y\left(y-2\right)}{\left(y+1\right)}\right)=\frac{y}{y-1}\)

a) \(P=\frac{\sqrt{x}}{\sqrt{x}-1}\)

b)\(x=6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\)

\(p=\frac{\left(\sqrt{5}-1\right)}{\sqrt{5}-2}=\left(\sqrt{5}-1\right)\left(\sqrt{5}+2\right)=3-\sqrt{5}\)

C)\(\frac{P}{\sqrt{x}}=\frac{1}{\sqrt{x}-1}\ge-1\) tuy nhiên đk: x khác 0=> dấu đẳng thức không xẩy ra (xem lại đề)

Đặt \(\sqrt{x}=y\\ \) ĐK tồn tại: hiển nhiên\(x\ge0\) và\(\left\{\begin{matrix}\sqrt{x}-2\ne0\\\sqrt{x}-1\ne0\\\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\ne0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\ne4\\x\ne1\\x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{\begin{matrix}y\ne2\\y\ne1\\y>0\end{matrix}\right.\)bạn chú ý cái đk thứ 3 nhé rất dẽ quên.

\(P=\left(\frac{y^2+3y+2}{\left(y-2\right)\left(y-1\right)}-\frac{y^2+y}{\left(y^2-1\right)}\right):\left(\frac{1}{y+1}+\frac{1}{y-1}\right)\)

\(P=\left(\frac{\left(y^2+3y+2\right)\left(y+1\right)}{\left(y-2\right)\left(y-1\right)\left(y+1\right)}-\frac{\left(y^2+y\right)\left(y-2\right)}{\left(y-2\right)\left(y-1\right)\left(y+1\right)}\right):\left(\frac{y-1+y+1}{\left(y+1\right)\left(y-1\right)}\right)\)

\(P=\left(\frac{\left(y+1\right)\left[\left(y+1\right)\left(y+2\right)-y\left(y-2\right)\right]}{\left(y-2\right)\left(y-1\right)}\right).\left(\frac{\left(y-1\right)\left(y+1\right)}{2y}\right)\)

\(P=\left(\frac{\left(y+1\right)\left(5y+2\right)}{\left(y-2\right)}\right).\left(\frac{\left(y+1\right)}{2y}\right)=\frac{\left(y+1\right)^2\left(5y+2\right)}{2y\left(y-2\right)}\)

sao không gọn đề sai chăng nghi con căn (x)-2 lắm

a) \(P=\frac{\left(\sqrt{x}+1\right)\left(5\sqrt{x}+2\right)}{2\sqrt{x}\left(\sqrt{x}-2\right)}\)

bạn trung học hay tiểu học vậy

a: \(P=\dfrac{-1+2\sqrt{x}-x+x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}:\dfrac{2x+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

b: Thay \(x=6-2\sqrt{5}\) vào P, ta được:

\(P=\dfrac{\sqrt{5}-1}{\sqrt{5}-2}=3+\sqrt{5}\)