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

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

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

⇔\(4x+3=0\) hay \(2x-1=0\)

⇔\(x=\dfrac{-3}{4}\) hay \(x=\dfrac{1}{2}\)

- Hic hic chắc mình làm mấy bài toán 8 cho khỏe :)

Câu 1:

a) A(x)+B(x)=x³+2x²-4x-4+x³+x²-6x-4=2x³+3x²-10x-8.

b) A(x)-B(x)=x³+2x²-4x-4-(x³+x²-6x-4)=x²+2x

1) 3x³y³    2)-4x³y³z   3)6x³y³   4) 3x⁴y³z²  5)  -5x²y⁷z²

6)12x⁹y¹¹z²