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