Question

Giari các phương trình 1. (2x2+3x-1)2-5(2x2+3x+3)+24=0

2. (x2-4)(x+3)=(x2-4)(x-1)

Answer 1

Bài 1:

Đặt \(t=2x^2+3x-1\) ta có:

\(t^2-5\left(t+4\right)+24=0\)

\(\Rightarrow t^2-5t-20+24=0\)

\(\Rightarrow t^2-5t+4=0\)

\(\Rightarrow\left(t-4\right)\left(t-1\right)=0\)\(\Rightarrow\left[\begin{matrix}t=4\\t=1\end{matrix}\right.\)

*)Xét \(2x^2+3x-1=4\)

\(\Rightarrow\left(x-1\right)\left(2x+5\right)=0\)\(\Rightarrow\left[\begin{matrix}x=1\\x=-\frac{5}{2}\end{matrix}\right.\)

*)Xét \(2x^2+3x-1=1\)

\(\Rightarrow\left(x+2\right)\left(2x-1\right)=0\)\(\Rightarrow\left[\begin{matrix}x=-2\\x=\frac{1}{2}\end{matrix}\right.\)

Answer 2

Bài 2:

\(\left(x^2-4\right)\left(x+3\right)=\left(x^2-4\right)\left(x-1\right)\)

\(\Rightarrow\left(x^2-4\right)\left(x+3\right)-\left(x^2-4\right)\left(x-1\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left[x+3-\left(x-1\right)\right]=0\)

\(\Rightarrow4\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)\(\Rightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)