mong mọi người giải giúp em vs 

ĐK: \(x\ge1\)

Đặt\(\left\{{}\begin{matrix}\sqrt{x+4}=a\\\sqrt{x-1}=b\end{matrix}\right.\)\(\left(a\ge\sqrt{5},b\ge0\right)\)

\(\Rightarrow a^2-b^2=5\)\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=5\)(1)

Mặt khác,\(PT\Leftrightarrow\)\(\left(a-b\right)\left(ab+1\right)=5\)(2)

Lấy \(\left(2\right)-\left(1\right)\Rightarrow\) \(\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=b\\a=1\left(l\right)\\b=1\left(tm\right)\end{matrix}\right.\)

Đến đây không biết giải tiếp, anh lo nhé :D

ĐKXĐ: \(-4\le x\le1\)

Đặt \(\sqrt{x+4}-\sqrt{1-x}=t\)

\(\Rightarrow t^2=5-2\sqrt{\left(x+4\right)\left(1-x\right)}\Rightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=\frac{5-t^2}{2}\)

Pt trở thành:

\(t\left(1+\frac{5-t^2}{2}\right)=3\Leftrightarrow t\left(7-t^2\right)=6\)

\(\Leftrightarrow t^3-7t+6=0\Leftrightarrow\left(t+3\right)\left(t-1\right)\left(t-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=-3\\t=1\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x+4}-\sqrt{1-x}=-3\\\sqrt{x+4}-\sqrt{1-x}=1\\\sqrt{x+4}-\sqrt{1-x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+4}+3=\sqrt{1-x}\left(vn\right)\\\sqrt{x+4}=1+\sqrt{1-x}\\\sqrt{x+4}=2+\sqrt{1-x}\end{matrix}\right.\) (1 vô nghiệm do \(VT\ge3;VP\le\sqrt{5}< 3\))

\(\Leftrightarrow\left[{}\begin{matrix}x+4=2-x+2\sqrt{1-x}\\x+4=5-x+4\sqrt{1-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{1-x}\left(x\ge-1\right)\\2x-1=4\sqrt{1-x}\left(x\ge\frac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=1-x\\4x^2-4x+1=16-16x\end{matrix}\right.\) \(\Leftrightarrow...\)