Question

(2x²+3x-1)²-5(2x²+3x+3)+24=0

tính sao anh em

 

Answer 1

Đặt \(2x^2+3x+1=y\).Ta có:

\(\left(y-2\right)^2-5\left(y+2\right)+24=0\)

\(\Leftrightarrow y^2-4y+4-5y-10+24=0\)

\(\Leftrightarrow y^2-9y+18=0\)

\(\Leftrightarrow\left(y-3\right)\left(y-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=3\\y=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+3x+1=3\\2x^2+3x+1=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)\left(x+2\right)=0\\\left(2x+5\right)\left(x-1\right)=0\end{cases}}\)

Vậy PT có 4 nghiệm là:\(\frac{1}{2}\)\(,\)\(-2,-\frac{5}{2},1\)