Question

Giải PT

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

Answer 1

Đkxđ x\(\ge\frac{1}{2}\) Đặt \(\left\{{}\begin{matrix}\sqrt{3x+1}=a\\\sqrt{2x-1}b\end{matrix}\right.\)>=0

\(\Leftrightarrow a^2-b^2=x+2\)

Khi đó

a + b= a² - b²

^{\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=a+b\)}

^{\(\Leftrightarrow\left(a+b\right)\left(a-b-1\right)=0\)}

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

^{\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{1}{2}\end{matrix}\right.\\\left[{}\begin{matrix}X=5\left(TM\right)\\X=1\left(TM\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow VL\)}