Question

(3x+2)(x²-1)= (9x²-4)(x+1)

Answer 1

\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)=\left(3x+2\right)\left(3x-2\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\)
\(\Rightarrow\)\(3x+2=0\)
hoặc \(x+1=0\)
hoặc \(1-2x=0\)
\(\Rightarrow x=-\frac{2}{3}\);   \(x=-1\);   \(x=\frac{1}{2}\)

Answer 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

Answer 3

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=\(\frac{-3}{2}\)

Answer 4

tớ viết nhầm dấu rồi!

xin lỗi!

Answer 5

(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 .

Answer 6

câu trả lời đúng nhất là 1+1=2 =))