ĐK: \(x\le\dfrac{3}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3-x}=a\\\sqrt{3-2x}=b\end{matrix}\right.\left(a>0,b\ge0\right)\)
\(pt\Leftrightarrow a-b=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\)
TH1: \(a=b\Leftrightarrow\sqrt{3-x}=\sqrt{3-2x}\Leftrightarrow x=0\left(tm\right)\)
TH2: \(a+b=1\Leftrightarrow\sqrt{3-x}+\sqrt{3-2x}=1\)
\(\Leftrightarrow6-3x+2\sqrt{\left(3-x\right)\left(3-2x\right)}=1\)
\(\Leftrightarrow2\sqrt{2x^2-9x+9}=3x-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(2x^2-9x+9\right)=\left(3x-5\right)^2\\3x-5\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+6x-11=0\\x\ge\dfrac{5}{3}\end{matrix}\right.\)
\(\Rightarrow\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=0\)
