Question

Giúp mình với những bài phân tích đa thức thành nhân tử với. ( dựa vào hằng đẳng thức)

(x-2)³-1

(x+3y)²-9y²

(x+3)²-(x-1)^2

Answer 1

\(\left(x-2\right)^3-1=\left(x-2\right)\left[\left(x-3\right)^2+x-2\right]=\left(x-2\right)\left(x^2+5x+7\right)\)

\(\left(x+3y\right)^2-9y^2=x\left(x+6y\right)\)

\(\left(x+3\right)^2-\left(x-1\right)^2=4\left(2x+4\right)=8\left(x+2\right)\)

Answer 2

a) \(\left(x-2\right)^3-1=\left(x-2\right)^3-1^3=\left(x-2-1\right)\left[\left(x-2\right)^2+\left(x-2\right)\cdot1+1^2\right]\)\(=\left(x-3\right)\left(x^2-4x+4+x-2+1\right)\)

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

b) \(\left(x+3y\right)^2-9y^2\)

\(=\left(x+3y\right)^2-\left(3y\right)^2\)

\(=\left(x+3y+3y\right)\left(x+3y-3y\right)\)

\(=x\left(x+6y\right)\)

c) \(\left(x+3\right)^2-\left(x-1\right)^2\)

\(=\left(x+3-x+1\right)\left(x+3+x-1\right)\)

\(=4\left(2x+2\right)\)

\(=8\left(x+1\right)\)