Em ko chắc đâu!
ĐK: chắc là x thuộc R:v
PT \(\Leftrightarrow\left(2x-1\right)\sqrt{x^2+2}+\left(2x+3\right)\sqrt{x^2+2x+3}+4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(\sqrt{x^2+2}-\frac{3}{2}\right)+10x+5+\left(2x+3\right)\left(\sqrt{x^2+2x+3}-\frac{3}{2}\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(\frac{x^2-\frac{1}{4}}{\sqrt{x^2+2}+\frac{3}{2}}\right)+10\left(x+\frac{1}{2}\right)+\left(2x+3\right)\left(\frac{x^2+2x+\frac{3}{4}}{\sqrt{x^2+2x+3}+\frac{3}{2}}\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(\frac{\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)}{\sqrt{x^2+2}+\frac{3}{2}}\right)+10\left(x+\frac{1}{2}\right)+\left(2x+3\right)\left(\frac{\left(x+\frac{1}{2}\right)\left(x+\frac{3}{2}\right)}{\sqrt{x^2+2x+3}+\frac{3}{2}}\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)\left[\frac{\left(2x-1\right)\left(x-\frac{1}{2}\right)}{\sqrt{x^2+2}+\frac{3}{2}}+10+\frac{\left(2x+3\right)\left(x+\frac{3}{2}\right)}{\sqrt{x^2+2x+3}}\right]=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)\left[\frac{2x^2-2x+\frac{1}{2}}{\sqrt{x^2+2}+\frac{3}{2}}+10+\frac{2x^2+6x+\frac{9}{2}}{\sqrt{x^2+2x+3}}\right]=0\)
Dễ thấy cái ngoặc to vô nghiệm suy ra \(x=-\frac{1}{2}\)
