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1) x2 -7x + 10 = x2 - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)
2) x2 + 3x + 2 = x2 + 2x + x + 2 = x(x + 2) + (x + 2) = (x + 1)(x + 2)
3) x2 - 7x + 12 = x2 - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)
4) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)
5) 16x - 5x2 - 3 = 15x - 5x2 + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x)
6) 6x2 + 7x - 3 = 6x2 - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)
7) 3x2 - 3x - 6 = 3x2 - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)
8) 3x2 + 3x - 6 = 3x2 - 3x + 6x - 6 = 3x(x - 1) + 6(x - 1) = (x - 1)(3x + 6) = 3(x - 1)(x + 2)
9) 6x2 - 13x + 6 = 6x2 - 9x - 4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3)
10) 6x2 + 15x + 6 = 6x2 + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)
11) 6x2 - 20x + 6 = 6x2 - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)
12) 8x2 + 5x - 3 = 8x2 + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)
Bài 2:
a: =x(x^2-25)
=x(x-5)(x+5)
b: =x(x-2y)+3(x-2y)
=(x-2y)(x+3)
c: =(2x-3)(4x^2+6x+9)+2x(2x-3)
=(2x-3)(4x^2+8x+9)
a) x3-10x2+21x
= x3-7x2-3x2+21x
= x2(x-7)-3x(x-7)
= (x2-3x)(x-7)
b) 3x3-7x2-20x
= x(3x2-7x-20)
= x(3x2+5x-12x-20)
= x[x(3x+5)-4(3x+5)]
= x(x-4)(3x+5)
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: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
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: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
Bài `1:`
`a)3x^3+6x^2=3x^2(x+2)`
`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`
Bài `2:`
`a)(2x-1)^2-25=0`
`<=>(2x-1-5)(2x-1+5)=0`
`<=>(2x-6)(2x+4)=0`
`<=>[(x=3),(x=-2):}`
`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`
`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`
`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`
`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`
1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)