Ta có: \(\sqrt{2x^2+7x+10}+\sqrt{2x^2+x+4}=3x+3\)
\(\Rightarrow\sqrt{2x^2+7x+10}+\sqrt{2x^2+x+4}-3x-3=0\)
\(\Rightarrow\sqrt{2x^2+7x+10}-7+\sqrt{2x^2+x+4}-5-3x+9=0\)
\(\Rightarrow\frac{2x^2+7x+10-49}{\sqrt{2x^2+7x+10}+7}+\frac{2x^2+x+4-25}{\sqrt{2x^2+x+4}+5}-3\left(x-3\right)=0\)
\(\Rightarrow\frac{\left(x-3\right)\left(2x+13\right)}{\sqrt{2x^2+7x+10}+7}+\frac{\left(x-3\right)\left(2x+7\right)}{\sqrt{2x^2+x+4}+5}-3\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(\frac{2x+13}{\sqrt{2x^2+7x+10}+7}+\frac{2x+7}{\sqrt{2x^2+x+4}+5}-3\right)=0\)
mà \(\frac{2x+13}{\sqrt{2x^2+7x+10}+7}+\frac{2x+7}{\sqrt{2x^2+x+4}}-3< 0\)
=> x - 3 = 0 => x = 3
Vậy x = 3
dân dương ơi bài này dễ mà nhân liên hợp là ok
