Question

Giải phương trình: \(3\left(x^2+2\right)=10\sqrt{x^3+1}\)

Answer 1

ĐKXĐ: ...

\(\Leftrightarrow3\left(x^2+2\right)=10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow3\left(a^2+b^2\right)=10ab\)

\(\Leftrightarrow3a^2-10ab+3b^2=0\)

\(\Leftrightarrow\left(a-3b\right)\left(3a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+1}=3\sqrt{x+1}\\3\sqrt{x^2-x+1}=\sqrt{x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=9x+9\\9x^2-9x+9=x+1\end{matrix}\right.\) \(\Leftrightarrow...\)