\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)=\left(3x+2\right)\left(3x-2\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\)
\(\Rightarrow\)\(3x+2=0\)
hoặc \(x+1=0\)
hoặc \(1-2x=0\)
\(\Rightarrow x=-\frac{2}{3}\); \(x=-1\); \(x=\frac{1}{2}\)
=> (3x+2) (x-1) (x+1) = (3x-2) (3x+2) (x+1)
=> (3x+2) (x-1) (x+1) - (3x-2) (3x+2) (x+1) = 0
=> (3x+2) (x+1) (x-1 - 3x +2) = 0
Tới đây thì cậu chia ra 3 trường hợp = 0 nhé
có:(3x+2)(x2-1)=(9x2-4)(x+1)
(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)
\(\Rightarrow\)(3x+2)(x+1)(3x+2-x+1)=0
(3x+2)(x+1)(2x+3)=0
đến đáy sẽ có 3 nghiệm là: x=\(\frac{-2}{3}\); x=-1; x=\(\frac{-3}{2}\)
(3x+2)(x^2-1)=(9x^2-4)(x+1)
(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)
(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0
[(3x+2)(x+1)][(x-1)-(3x-2)]=0
(3x^2+3x+2x+2)(x-1-3x+2)=0
(3x^2+5x+2)(1-2x)=0
Ma : 3x^2+5x+2=0
Hay : 3x^2+3x+2x+2=0
3x(x+1)+2(x+1)=0
(x+1)(3x+2)=0
\(\Rightarrow\)(x+1)(3x+2)(1-2x)=0
TH1 : x+1=0=>x=-1
TH2 : 3x+2=0=>x=-2/3
TH3 : 1-2x=0=>x=1/2
k mk nha .
câu trả lời đúng nhất là 1+1=2 =))
