Question

[ngọc hân]

Bài 1: Phân tích đa thức thành nhân tử

a) (6x+3)-(2x-5)(2x+1)

b) (3x-2)(4x-3)-(2-3x)(x-1)-2(3x-2)(x+1)

Bài 2^*:Phân tích đa thức thành nhân tử

a) (a-b)(a+2b)-(b-a)(2a-b)-(a-b)(a+3b)

b) 5xy³-2xy²-15y²+6z

c) (x+y)(2x-y)+(2x-y)(3x-y)-(y-2x)

d) ab³c²-a²b²c²+ab²c³-a²bc

e) x²(y-z)+y²(z-x)+z²(x-y)

f) x²-6xy+9y²+4x-12y

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

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

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

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

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

\(=\left(3x-2\right)\left(3x-6\right)\)

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

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

\(=\left(x-3y\right)\left(x-3y+4\right)\)