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![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3x2 – 7x + 2
\(=3x^2-6x-x+2\)
\(=\left(3x^2-6x\right)-\left(x-2\right)\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b) a(x2 + 1) – x(a2 + 1)
\(=ax^2+a-\left(a^2x+x\right)\)
\(=a\left(x^2+1\right)-x\left(a^2+1\right)\)
.......?
a) Ta có: \(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)
\(=x^2a+a-a^2x-x\)
\(=\left(x^2a-a^2x\right)+\left(a-x\right)\)
\(=xa\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(xa-1\right)\)
c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)
\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)
\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)
\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)
\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)
\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)
\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2a-3\right)\left(a+1\right)-\left(a^2+6a+9\right):\left(a+3\right)\)
\(=\left(2a^2+2a-3a-3\right)-\left(a+3\right)^2:\left(a+3\right)\)
\(=2a^2-a-3-\left(a+3\right)\)
\(=2a^2-a-3-a-3\)
\(=2a^2-2a-6\)
b) \(\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\)
\(=\left(3x-5y\right)\cdot x^2y^2-3x^2y^2+4x^2y^3\)
\(=3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\)
\(=3x^3y^2-x^2y^3-3x^2y^2\)
c) \(x\left(x-2\right)^2-\left(x+2\right)\left(x^2-2x+4\right)+4x^2\)
\(=x\left(x^2-4x+4\right)-\left(x^3+8\right)+4x^2\)
\(=x^3-4x^2+4x-x^3-8+4x^2\)
\(=\left(x^3-x^3\right)+\left(-4x^2+4x^2\right)+4x-8\)
\(=4x-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(2a-3\right)\left(a+1\right)+\left(a^2+6a+9\right):\left(a+3\right)\\ =2a^2-a-3+\left(a+3\right)^2:\left(a+3\right)\\ =2a^2-a-3+a+3\\ =2a^2\\ b,\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\\ =3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\\ =3x^3y^2-3x^2y^2-x^2y^3\\ c,x\left(x-2\right)^2-\left(x+2\right)\left(x^2-2x+4\right)+4x^2\\ =x^3-4x^2+4x-x^3-8+4x^2\\ =4x-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: Ta có: \(A=\left(k-4\right)\left(k^2+4k+16\right)-\left(k^3+128\right)\)
\(=k^3-64-k^3-128\)
=-192
b: Ta có: \(B=\left(2m+3n\right)\left(4m^2-6mn+9n^2\right)-\left(3m-2n\right)\left(9m^2+6mn+4n^2\right)\)
\(=8m^3+27n^3-27m^3+8n^3\)
\(=-19m^3+35n^3\)
Bài 4:
a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)
\(\Leftrightarrow9x=9\)
hay x=1
b: ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Leftrightarrow x^3+8-x^3+2x=15\)
\(\Leftrightarrow2x=7\)
hay \(x=\dfrac{7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)
\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
hay P=x+3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 12:
1) A = x2 - 6x + 11
= (x2 - 6x + 9) + 2
= (x - 3)2 + 2
Ta có: (x - 3)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 3 = 0 ⇔ x = 3
Do đó: (x - 3)2 + 2 ≥ 2
Hay A ≥ 2
Dấu ''='' xảy ra khi x = 3
Vậy Min A = 2 tại x = 3
2) B = x2 - 20x + 101
= (x2 - 20x + 100) + 1
= (x - 10)2 + 1
Ta có: (x - 10)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 10 = 0 ⇔ x = 10
Do đó: (x - 10)2 + 1 ≥ 1
Hay B ≥ 1
Dấu ''='' xảy ra khi x = 10
Vậy Min B = 1 tại x = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)