\(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\)(ĐKXĐ: \(\frac{3}{2}\le x\le\frac{5}{2}\))
\(\Leftrightarrow2\sqrt{2x-3}+2\sqrt{5-2x}=6x^2-24x+28\)
\(\Leftrightarrow6x^2-24x+28-2\sqrt{2x-3}-2\sqrt{5-2x}=0\)
\(\Leftrightarrow\left(2x-3-2\sqrt{2x-3}+1\right)+\left(5-2x-2\sqrt{5-2x}+1\right)+6x^2-24x+24=0\)
\(\Leftrightarrow\left(\sqrt{2x-3}-1\right)^2+\left(\sqrt{5-2x}-1\right)^2+6\left(x-2\right)^2=0\)
Do \(\left(\sqrt{2x-3}-1\right)^2\ge0;\left(\sqrt{5-2x}-1\right)^2\ge0;6\left(x-2\right)^2\ge0\forall x\in R\)
Nên \(\hept{\begin{cases}\sqrt{2x-3}-1=0\\\sqrt{5-2x}-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-3=1\\5-2x=1\\x=2\end{cases}}\Leftrightarrow x=2\)(t/m ĐKXĐ)
Vậy pt có nghiệm duy nhất là x=2.
