\(\sqrt{x^2+3x-2}=\sqrt{1+x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}1+x\ge0\\x^2+3x-2=1+x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=1\\x=-3\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x=1\)

Em trục căn thức:

\(\sqrt{x+3}-2\sqrt{x}=\sqrt{2x+2}-\sqrt{3x+1}\)

<=> \(\frac{-3x+3}{\sqrt{x+3}+2\sqrt{x}}=\frac{-x+1}{\sqrt{2x+2}+\sqrt{3x+1}}\)

=> nhân tử chung là -x + 1 . Tự làm tiếp nhé!

làm như cô thì vẫn cần phải đánh giá rất khó chịu nhé

\(\sqrt{x+3}-2\sqrt{x}=\sqrt{2x+2}-\sqrt{3x+1}\left(ĐKXĐ:x\ge0\right)\)

\(< =>\sqrt{x+3}-\sqrt{2x+2}+\sqrt{3x+1}-2\sqrt{x}=0\)

\(< =>\frac{\sqrt{x+3}^2-\sqrt{2x+2}^2}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{\sqrt{3x+1}^2-4\sqrt{x}^2}{\sqrt{3x+1}+2\sqrt{x}}=0\)

\(< =>\frac{x+3-2x-2}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{3x+1-4x}{\sqrt{3x+1}+2\sqrt{x}}=0\)

\(< =>\frac{1-x}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{1-x}{\sqrt{3x+1}+2\sqrt{x}}=0\)

\(< =>\left(1-x\right)\left(\frac{1}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{1}{\sqrt{3x+1}+2\sqrt{x}}\right)=0< =>x=1\)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-3\end{matrix}\right.\)

\(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\)

\(\Leftrightarrow-x^2-3x+10=3\sqrt{x^2+3x}\)

\(\Leftrightarrow x^2+3x+3\sqrt{x^2+3x}-10=0\)

Đặt \(\sqrt{x^2+3x}=t\ge0\)

\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+3x}=2\)

\(\Leftrightarrow x^2+3x=4\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)