ĐKXĐ: \(x\in R\)
\(3x^2-5x+6=2x\cdot\sqrt{x^2-x+2}\)
=>\(3x^2-6x+x-2+8=2\cdot\sqrt{x^4-x^3+2x^2}\)
=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\left(\sqrt{x^4-x^3+2x^2}-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-x^3+2x^2-16}{\sqrt{x^4-x^3+2x^2}+4}\)
=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-2x^3+x^3-2x^2+4x^2-8x+8x-16}{\sqrt{x^4-x^3+2x^2}+4}\)
=>\(\left(x-2\right)\left(3x+1\right)=\dfrac{2\left(x-2\right)\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\)
=>\(\left(x-2\right)\left[\left(3x+1\right)-\dfrac{2\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\right]=0\)
=>x-2=0
=>x=2(nhận)
\(3x^2-5x+6=2x\sqrt{x^2-x+2}\)
\(\Leftrightarrow\left[x^2-2x\sqrt{x^2-x+2}+\left(x^2-x+2\right)\right]+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x^2-x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{x^2-x+2}\\x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
Thử lại ta thấy nghiệm \(x=2\) thỏa phương trình ban đầu.
