Bài này phải giải phương trình bậc 2 mà ghi là lớp 8 cũng khó!
\(\left(2x^2+3x-1\right)^2-5\left(2x^2+3x-3\right)-24=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)^2-5\left(2x^2+3x-1\right)-14=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)^2-7\left(2x^2+3x-1\right)+2\left(2x^2+3x-1\right)-14=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(2x^2+3x-1-7\right)+2\left(2x^2+3x-1-7\right)=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(2x^2+3x-8\right)+2\left(2x^2+3x-8\right)=0\)
\(\Leftrightarrow\left(2x^2+3x-1+2\right)\left(2x^2+3x-8\right)=0\)
\(\Leftrightarrow\left(2x^2+3x+1\right)\left(2x^2+3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x+1=0\\2x^2+3x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\\x=\dfrac{-3+\sqrt{73}}{4}\\x=\dfrac{-3-\sqrt{73}}{4}\end{matrix}\right.\)
pt có 4 nghiệm.
