Question

giải phương trình
\(\sqrt{3x^2-2x+15}+\sqrt{3x^2-2x+8}=7\)

Answer 1

Đặt \(\left\{{}\begin{matrix}\sqrt{3x^2-2x+15}=a>0\\\sqrt{3x^2-2x+8}=b>0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=7\)

Pt trở thành:

\(a+b=a^2-b^2\)

\(\Leftrightarrow a+b=\left(a-b\right)\left(a+b\right)\)

\(\Rightarrow a-b=1\Rightarrow a=b+1\)

\(\Rightarrow\sqrt{3x^2-2x+15}=\sqrt{3x^2-2x+8}+1\)

\(\Leftrightarrow3x^2-2x+15=3x^2-2x+9+2\sqrt{3x^2-2x+8}\)

\(\Leftrightarrow\sqrt{3x^2-2x+8}=3\)

\(\Leftrightarrow3x^2-2x-1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)