a) \(x^2+3-\sqrt{2x^2-3x+2}=\frac{3}{2}\left(x+1\right)\)
\(\Leftrightarrow x^2.2+3.2-\sqrt{2x^2-3x+2}.3=\frac{3}{2}\left(x+1\right).2\)
\(\Leftrightarrow2x^2+6-\sqrt{2x^2-3x+2}=3\left(x+1\right)\)
\(\Leftrightarrow2x^2+6-2\sqrt{2x^2-3x+2}=3x+3\)
\(\Leftrightarrow-2\sqrt{2x^2-3x+2}+6=3x^2+3-2x^2\)
\(\Leftrightarrow-2\sqrt{2x^2-3x+2}=3x+3-2x^2-6\)
\(\Leftrightarrow-2\sqrt{2x^2-3x+2}=-2x^3+3x-3\)
\(\Leftrightarrow\left(-2\sqrt{2x^2-3x+2}\right)^2=\left(-2x^2+3x-3\right)^2\)
\(\Leftrightarrow8x^2-12x+8=4x^4-12x^3+21x^2-18x+9\)
\(\Leftrightarrow4x^2-12x^3+12x^2-6x+1=0\)
\(\Leftrightarrow\left(x-2\right)^2\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: nghiệm phương trình là \(\left\{1;\frac{1}{2}\right\}\)
b) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Xét \(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)
\(=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=\left|1\right|=1\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(3-\sqrt{x-1}\right)\ge0\Leftrightarrow5\le x\le10\)
