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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.
$3x(x-2)-x+2=3x(x-2)-(x-2)=(x-2)(3x-1)$
\(a,ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\\ =a^2b-ab^2+b^2c-bc^2+ca\left(c-a\right)\\ =\left(a^2b-bc^2\right)-\left(ab^2-b^2c\right)+ca\left(c-a\right)\\ =b\left(a-c\right)\left(a+c\right)-b^2\left(a-c\right)-ca\left(a-c\right)\\ =\left(a-c\right)\left(ab+bc-b^2-ca\right)\\ =\left(a-c\right)\left(b-c\right)\left(a-b\right)\)
\(b,x^2-3xy-10y^2\\ =x^2+2xy-5xy-10y^2\\ =x\left(x+2y\right)-5y\left(x+2y\right)=\left(x-5y\right)\left(x+2y\right)\)
\(c,3x\left(x-2\right)-x+2=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)
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a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
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a) \(4x^2-1=\left(2x+1\right)\left(2x-1\right)\)
b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)
c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)
\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)
\(=3b\left(2a-b\right)\)
`a, 4x^2-1 = (2x+1)(2x-1)`
`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`
`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`
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Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
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) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
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a(b^2 +c^2 + bc) + b(c^2 + a^2 +ac) + c(a^2 + b^2 + ab)
= a.b^2 + a.c^2 + b.c^2 + b.a^2 + c.a^2 + c.b^2 + 3abc
= (a.b^2 + b.a^2 +abc) + ( a.c^2 + c.a^2 + abc) + (c.b^2 + b.c^2 + abc)
= ab(a+b+c) + ac(a +b +c) + bc(a+b+c)
=(a+b+c)(ab+ac+bc)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=4x\left(x^2-2x+1\right)=4x\left(x-1\right)^2\\ B=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\\ C=\left(x-2\right)\left(x^2+2x+4\right)+3\left(x-2\right)=\left(x-2\right)\left(x^2+2x+7\right)\)
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) \(C=\left(x-2\right)\left(x^2+2x+4\right)+3\left(x-2\right)=\left(x-2\right)\left(x^2+2x+7\right)\)
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