\(\left(5x^2-2x+19\right)^2=\left(3x^2+10x-8\right)^2\)
\(\Leftrightarrow\left(5x^2-2x+19\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+19-3x^2-10x+8\right)\left(5x^2-2x+19+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-12x+27\right)\left(8x^2+8x+11\right)=0\)
....
\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)
\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right).\left(5x^2-2x+10+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-12x+18\right).\left(8x^2+8x+2\right)=0\)
\(\Leftrightarrow2.\left(x^2-6x+9\right).2.\left(4x^2+4x+1\right)=0\)
\(\Leftrightarrow2.\left(x-3\right)^2.2.\left(2x+1\right)^2=0\)
\(\Leftrightarrow4.\left(x-3\right)^2.\left(2x+1\right)^2=0\)
Vì \(4\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{3;-\frac{1}{2}\right\}.\)
Chúc bạn học tốt!
