ĐK \(-3\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow2\sqrt{x+3}-4-2\sqrt{3-2x}+2+2x^2+3x-5=0\)
\(\Leftrightarrow2\left(\sqrt{x+3}-2\right)-2\left(\sqrt{3-2x}-1\right)+\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\dfrac{2\left(\sqrt{x+3}-2\right)\left(\sqrt{x+3}+2\right)}{\sqrt{x+3}+2}-\dfrac{2\left(\sqrt{3-2x}-1\right)\left(\sqrt{3-2x}+1\right)}{\sqrt{3-2x}+1}+\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\dfrac{2\left(x-1\right)}{\sqrt{x+3}+2}-\dfrac{2\left(2-2x\right)}{\sqrt{3-2x}+1}+\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\dfrac{2\left(x-1\right)}{\sqrt{x+3}+2}+\dfrac{4\left(x-1\right)}{\sqrt{3-2x}+1}+\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{2}{\sqrt{x+3}+2}+\dfrac{4}{\sqrt{3-2x}+1}+2x+5\right)=0\)
ta thấy \(\dfrac{2}{\sqrt{x+3}+2}+\dfrac{4}{\sqrt{3-2x}+1}+2x+5\ne0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
