Question

giải các phươngtrình

1. x³+3x²+x+3 = 0

2. (x²+x+1)(x²+x+2) = 2

3. (2x²+3x-1)²-5(2x²+3x+3)+24=0

4 (x²-4)(x+3)=(x²-4)(x-1)

Answer 1

1. Ta có \(x^3+3x^2+x+3=0\)

\(\Leftrightarrow\left(x^3+3x^2\right)+\left(x+3\right)=0\)

\(\Leftrightarrow x^2\left(x+3\right)+\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)

Nếu x+3=0 =>x=-3

Nếu \(x^2+1=0\) =>x\(=\varnothing\) (vì \(x^2+1>0\))

Vậy x=-3

Answer 2

2) đặt x^2+x+1 = t

=> x^2 +x +2 =t+1

pt => t(t+1)=2

t^2 + t -2 =0

\(\Rightarrow\left[\begin{matrix}t=1\\t=-2\end{matrix}\right.\)

voi t=1 => x^2 +x+1=1

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

voi t=-2 => x^2+x+1=-2

=> x^2+x+3=0(vo nghiem)

cau 3 lam nhu cau 2

4) pt <=> (x^2-4)(x+3-x+1)=0

ban tu giai not nha