Câu hỏi của Anh Tuấn Lê

Question

Bài 1. Phân tích các đa thức sau thành nhân tử:

a) a^2-b^2-2a+2b

b) 3x-3y-5x(y-x)

c) (x-y+4)^2 - (2x+3y-1)^2

d)16-x^2+4xy-4y^2

e)(x+3)^3 + (x-3)^3

f) x^4 + x^3 + 2x^2 +x +1

g) 9x^2- 3xy+y-6x+1...

Nguyễn Lê Phước Thịnh · Accepted Answer

Bài 1:

a) Ta có: $a^2-b^2-2a+2b$

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

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

b) Ta có: $3x-3y-5x\left(y-x\right)$

$=3\left(x-y\right)+5x\left(x-y\right)$

$=\left(x-y\right)\left(3+5x\right)$

c) Ta có: $\left(x-y+4\right)^2-\left(2x+3y-1\right)^2$

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

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

d) Ta có: $16-x^2+4xy-4y^2$

$=16-\left(x^2-4xy+4y^2\right)$

$=16-\left(x-2y\right)^2$

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

e) Ta có: $\left(x+3\right)^3+\left(x-3\right)^3$

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

$=2x\cdot\left(x^2+6x+9-x^2+9+x^2-6x+9\right)$

$=2x\cdot\left(x^2+27\right)$

f) Ta có: $x^4+x^3+2x^2+x+1$

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

$=\left(x^2+1\right)^2+x\left(x^2+1\right)$

$=\left(x^2+1\right)\left(x^2+1+x\right)$

g) Ta có: $9x^2-3xy+y-6x+1$

$=\left(9x^2-6x+1\right)-\left(3xy-y\right)$

$=\left(3x-1\right)^2-y\left(3x-1\right)$

$=\left(3x-1\right)\left(3x-1-y\right)$

h) Ta có: $x^3-4x^2+12x-27$

$=\left(x^3-27\right)-4x\left(x-3\right)$

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

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

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

Bài 2:

Ta có: $x^3+x^2z+y^2z-xyz+y^3$

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

$=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)$

$=\left(x^2-xy+y^2\right)\left(x+y+z\right)$

$=0\cdot\left(x^2-xy+y^2\right)=0$(đpcm)Câu trả lời: Bài 1: