Đặt \(\sqrt{2x^2+5x+12}=a\text{ và }\sqrt{2x^2+3x+2}=b\left(a\text{ và }b\ge0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=x+5\left(\text{✳}\right)\\a^2-b^2=2\left(x+5\right)\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)\left(a+b\right)=2\left(a+b\right)\)
\(\Rightarrow a=b+2\text{. Thay vào }\left(\text{✳}\right)\)
\(\Rightarrow\left(b+2\right)+b=x+5\)
\(\Leftrightarrow b=\dfrac{x+3}{2}\)
\(\Rightarrow2\sqrt{2x^2+3x+2}=x+3\)
\(\Leftrightarrow8x^2+12x+8=x^2+6x+9\)
\(\Leftrightarrow\left(7x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-1\end{matrix}\right.\)
☠ Bạn tự kết luận nha >..<"
