Question

giải phương trình sau 

\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

Answer 1

\(\Rightarrow\left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\)

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

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

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

=> 3x + 2 = 0 => x = -2/3

hoặc x + 1 = 0 => x = -1

hoặc 1 - 2x = 0 => x = 1/2

Answer 2

(3x+2)(x²-1) = (9x²-4)(x+1) => (3x+2)(x-1)(x+1) = [ (3x)²- 2² ](x+1) => (3x+2)(x-1) = (3x+2)(3x-2)

=> x-1 = 3x-2 => x = 3x-1 => 1 = 3x-x = 2x => x = 1:2 = 0,5