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29 tháng 8 2021

Trả lời:

a, \(27a^2b^2-18ab+3=3\left(9a^2b^2-6ab+1\right)=3\left(3ab-1\right)^2\)

b, \(x^2+2xy+y^2-xz-yz\)

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

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

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

c, \(a^4+a^3-a^2-a\)

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

\(=a^3\left(a+1\right)-a\left(a+1\right)\)

\(=a\left(a+1\right)\left(a^2-1\right)\)

\(=a\left(a+1\right)\left(a-1\right)\left(a+1\right)\)

\(=a\left(a+1\right)^2\left(a-1\right)\)

d, \(a^3-b^3+2b-2a\)

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

\(=\left(a-b\right)\left(a^2+ab+b^2\right)-2\left(a-b\right)\)

\(=\left(a-b\right)\left(a^2+ab+b^2-2\right)\)

29 tháng 8 2021

Trả lời:

\(27a^2b^2-18ab+3\)

\(=3\left(9a^2b^2-6ab+1\right)\)

\(=3\left[\left(3ab\right)^2-2.3ab.1+1^2\right]\)

\(=3\left(3ab-1\right)^2\)

a: 3x^2-12y^2

=3(x^2-4y^2)

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

b: 5xy^2-10xyz+5xz^2

=5x(y^2-2yz+z^2)

=5x(y-z)^2

g: (a+b+c)^3-a^3-b^3-c^3

=(a+b+c-a)[(a+b+c)^2+a(a+b+c)+a^2]-(b+c)(b^2-bc+c^2)

=(b+c)[a^2+b^2+c^2+2ab+2ac+2bc+a^2+ab+ac+a^2-b^2+bc-c^2]

=(b+c)[3a^2+3ab+3bc+3ac]

=3(a+b)(b+c)(a+c)

14 tháng 7 2016

다른 사람을 물어보세요! 알았지? 난 대답을 모르겠어요. 정말 미안 해요. 당신에게 좋은 날이 젠장!다른 사람을 물어보세요! 알았지? 난 대답을 모르겠어요. 정말 미안 해요. 당신에게 좋은 날이 젠장!

14 tháng 7 2016

câu nào mình biết mình trl trc nha:

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

\(=x^4+2x^2+1-4x^2\)

\(=x^4-2x^2+1\)

\(\left(x^2-1\right)^2\)

15 tháng 9 2016

4a2b2 + 36a2b3 + 6ab4

= 2ab2(2a + 18ab + 3b2)

3n(m - 3) + 5m(m - 3)

= (3n + 5m)(m - 3)

2a(x - y) - (y - x)

= (x - y)(2a + 1)

4a2b3 - 6a3b2

= 2a2b2(2b - 3a)

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: \(x^2-6xy+9y^2+4x-12y\)

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

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

NV
16 tháng 7 2021

a.

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

\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)

\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)

b.

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

\(=\left(x-y\right)\left(x-y-z\right)\)

c.

\(=x^4-1+4x^2-4\)

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

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

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

a) Ta có: \(x^3+3x^2+3x+1-27z^3\)

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

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

b) Ta có: \(x^2-2xy+y^2-zx+yz\)

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

\(=\left(x-y\right)\left(x-y-z\right)\)

c) Ta có: \(x^4+4x^2-5\)

\(=x^4+4x^2+4-9\)

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

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

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