x=-4, x=-5/2, x=-4/3, x=-1, x=0, x=1
bậc to quá nghĩ cách giải đã
Phan Đỗ Hoàng Linh Đăt \(a=x^2+3x-4,b=3x^2+7x+4\Rightarrow a+b=4x^2+10x\), ta có
\(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow a^3+b^3=a^3+b^3+3ab\left(a+b\right)\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a=-b\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x-4=0\\3x^2+7x+4=0\\x^2+3x-4=-\left(3x^2+7x+4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+4\right)=0\\\left(x+1\right)\left(3x+4\right)=0\\2x\left(2x+5\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\\x=-4\\x=-\dfrac{4}{3}\text{hoăc}x=-\dfrac{5}{2}\end{matrix}\right.\)