Question

\(\sqrt{x^2-3x+2}\left(x-3\right)=x^2-8x+15\)

Answer 1

ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x^2-3x+2}=x-5\left(1\right)\end{matrix}\right.\)

Xét (1) với \(x\ge5\)

\(\Leftrightarrow x^2-3x+2=\left(x-5\right)^2\)

\(\Leftrightarrow x^2-3x+2=x^2-10x+25\)

\(\Leftrightarrow7x=23\)

\(\Rightarrow x=\frac{23}{7}< 5\left(l\right)\)

Vậy pt có nghiệm duy nhất \(x=3\)