Lời giải:
ĐKXĐ:
\(x\geq \frac{1}{2}\)
Ta có: \((3x^2-6x)(\sqrt{2x-1}+1)=2x^3-5x^2+4x-4\)
\(\Leftrightarrow 3x(x-2)(\sqrt{2x-1}+1)=(2x^3-4x^2)-(x^2-4x+4)\)
\(\Leftrightarrow 3x(x-2)(\sqrt{2x-1}+1)=2x^2(x-2)-(x-2)^2=(x-2)(2x^2-x+2)\)
\(\Leftrightarrow (x-2)[3x(\sqrt{2x-1}+1)-(2x^2-x+2)]=0\)
\(\Rightarrow \left[\begin{matrix} x-2=0\rightarrow x=2\\ 3x(\sqrt{2x-1}+1)=2x^2-x+2(*)\end{matrix}\right.\)
Xét \((*)\)
\(\Leftrightarrow 3x\sqrt{2x-1}=2x^2-4x+2\)
\(\Leftrightarrow 3x\sqrt{8x-4}=4x^2-8x+4\)
\(\Leftrightarrow 3x(\sqrt{8x-4}-x)=x^2-8x+4\)
\(\Leftrightarrow 3x.\frac{8x-4-x^2}{\sqrt{8x-4}+x}=x^2-8x+4\)
\(\Leftrightarrow (x^2-8x+4)\left(1+\frac{3x}{\sqrt{8x-4}+x}\right)=0\)
Thấy rằng biểu thức trong ngoặc lớn luôn lớn hơn $0$ với mọi \(x\geq \frac{1}{2}\)
Do đó \(x^2-8x+4=0\Leftrightarrow x=4\pm 2\sqrt{3}\) (đều thỏa mãn)
Vậy..............
