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AH
Akai Haruma
Giáo viên
28 tháng 8 2021

Lời giải:

a.

$ab(a-b)+bc(b-c)+ca(c-a)$

$=ab(a-b)-bc[(a-b)+(c-a)]+ca(c-a)$

$=ab(a-b)-bc(a-b)-bc(c-a)+ca(c-a)$

$=(a-b)(ab-bc)-(c-a)(bc-ca)=b(a-b)(a-c)-c(c-a)(b-a)$

$=b(a-b)(a-c)-c(a-c)(a-b)=(a-b)(b-c)(a-c)$

b.

$x^2-3xy-10y^2=(x^2+2xy)-(5xy+10y^2)$

$=x(x+2y)-5y(x+2y)=(x+2y)(x-5y)$

c.

$3x(x-2)-x+2=3x(x-2)-(x-2)=(x-2)(3x-1)$

28 tháng 8 2021

\(a,ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\\ =a^2b-ab^2+b^2c-bc^2+ca\left(c-a\right)\\ =\left(a^2b-bc^2\right)-\left(ab^2-b^2c\right)+ca\left(c-a\right)\\ =b\left(a-c\right)\left(a+c\right)-b^2\left(a-c\right)-ca\left(a-c\right)\\ =\left(a-c\right)\left(ab+bc-b^2-ca\right)\\ =\left(a-c\right)\left(b-c\right)\left(a-b\right)\)

\(b,x^2-3xy-10y^2\\ =x^2+2xy-5xy-10y^2\\ =x\left(x+2y\right)-5y\left(x+2y\right)=\left(x-5y\right)\left(x+2y\right)\)

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

 

a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)

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

Bài 1: 

a: \(4a^2-6b=2\left(2a^2-3b\right)\)

b: \(m^3n-2m^2n^2-mn\)

\(=mn\left(m^2-2mn-1\right)\)

2 tháng 10 2021

Bài 1:

a) \(4a^2-6b=2\left(a^2-3b\right)\)

b) \(=mn\left(m^2-2mn-1\right)\)

Bài 2:

a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)

b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)

29 tháng 6 2017

a) (a+b+c)^2 + (a+b-c)^2 - 4c^2

\(=\left(a+b+c\right)^2+\left[\left(a+b-c\right)^2-\left(2c\right)^2\right]\)

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

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

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

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

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

b) 4a^2b^2 - (a^2+b^2-c^2)^2

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

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

\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)

\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)

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

\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)

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

\(=a^3\left(c-b\right)+\left(c-b\right)\left(bc^2+b^2c\right)-\left(c-b\right)\left(ac^2+abc+ab^2\right)\)

\(=\left(c-b\right)\left(a^3+bc^2+b^2c-ac^2-abc-ab^2\right)\)

29 tháng 6 2017

a) (a+b+c)^2 + (a+b-c)^2 - 4c^2

\(=\left(a+b+c\right)^2+\left[\left(a+b-c\right)^2-\left(2c\right)^2\right]\)

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

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

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

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

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

b) 4a^2b^2 - (a^2+b^2-c^2)^2

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

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

\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)

\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)

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

\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)

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

\(=a^3\left(c-b\right)+\left(c-b\right)\left(bc^2+b^2c\right)-\left(c-b\right)\left(ac^2+abc+ab^2\right)\)

\(=\left(c-b\right)\left(a^3+bc^2+b^2c-ac^2-abc-ab^2\right)\)

7 tháng 11 2021

\(A=4x\left(x^2-2x+1\right)=4x\left(x-1\right)^2\\ B=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\\ C=\left(x-2\right)\left(x^2+2x+4\right)+3\left(x-2\right)=\left(x-2\right)\left(x^2+2x+7\right)\)

7 tháng 11 2021

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

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

c) \(C=\left(x-2\right)\left(x^2+2x+4\right)+3\left(x-2\right)=\left(x-2\right)\left(x^2+2x+7\right)\)

15 tháng 7 2017

\(D=a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(D=a^3\left(b-c\right)+\left[b^3\left(c-a\right)+c^3\left(a-b\right)\right]\)

\(D=a^3\left(b-c\right)\left(b^3c-ab^3+ac^3-bc^3\right)\)

\(D=a^3\left(b-c\right)\left[\left(b^3c-bc^3\right)-\left(ab^3-ac^3\right)\right]\)

\(D=a^3\left(b-c\right)\left[bc\left(b^2-c^2\right)-a\left(b^3-c^3\right)\right]\)

\(D=a^3\left(b-c\right)\left[bc\left(b-c\right)\left(b+c\right)-a\left(b-c\right)\left(b^2+bc+c^2\right)\right]\)

\(D=\left(b-c\right)\left[a^3+bc\left(b+c\right)-a\left(b^2+bc+c^2\right)\right]\)

\(D=\left(b-c\right)\left(a^3+b^2c+bc^2-ab^2-abc-ac^2\right)\)

\(D=\left(b-c\right)\left[\left(b^2c-ab^2\right)+\left(bc^2-abc\right)-\left(ac^2-a^3\right)\right]\)

\(D=\left(b-c\right)\left[b^2\left(c-a\right)+bc\left(c-a\right)-a\left(c^2-a^2\right)\right]\)

\(D=\left(b-c\right)\left[b^2\left(c-a\right)+bc\left(c-a\right)-a\left(c-a\right)\left(c+a\right)\right]\)

\(D=\left(b-c\right)\left(c-a\right)\left[b^2+bc-a\left(c+a\right)\right]\)

\(D=\left(b-c\right)\left(c-a\right)\left(b^2+bc-ac-a^2\right)\)

\(D=\left(b-c\right)\left(c-a\right)\left[\left(bc-ac\right)+\left(b^2-a^2\right)\right]\)

\(D=\left(b-c\right)\left(c-a\right)\left[c\left(b-a\right)+\left(b-a\right)\left(b+a\right)\right]\)

\(D=\left(b-c\right)\left(c-a\right)\left(b-a\right)\left(c+b+a\right)\)

\(D=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)

Chúc bạn học tốt.