\(ĐKXĐ:\dfrac{74}{27}\le x\le\dfrac{10}{3}\)
PT đã cho tương đương với:
\(4-3\sqrt{10-3x}=x^2-4x+4\)
\(\Leftrightarrow x^2-4x+3+3\sqrt{10-3x}-3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-3\left(1-\sqrt{10-3x}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-3.\dfrac{3\left(x-3\right)}{1+\sqrt{10-3x}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1-\dfrac{9}{1+\sqrt{10-3x}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\left(1\right)\\\left(x-1\right)\left(1+\sqrt{10-3x}\right)=9\left(2\right)\end{matrix}\right.\)
Ta có:
\(pt\left(1\right)\Leftrightarrow x=3\left(tm\right)\)
\(pt\left(2\right):\left(x-1\right)\left(1+\sqrt{10-3x}\right)=9\)
mà \(\left(x-1\right)\left(1+\sqrt{10-3x}\right)\le\dfrac{7}{3}.\dfrac{7}{3}\) nên \(pt\left(2\right)\) vô nghiệm
Vậy pt đã cho có tập nghiệm \(S=\left\{3\right\}\)
