Phân tích đa thức thành nhân tử:
a(b+c)2+b(a+c)2+c(a+b)2-4abc
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a, \(4abc-8ab^2c=4abc\left(1-2b\right)\)
b, \(x^2\left(2a-1\right)+x\left(1-2a\right)=x^2\left(2a-1\right)-x\left(2a-1\right)\)
\(=x\left(x-1\right)\left(2a-1\right)\)
c, \(9a^4\left(a-2\right)+a^2\left(a-2\right)=a^2\left(9a^2+1\right)\left(a-2\right)\)
d, \(\left(a-4\right)\left(2a-1\right)-8a+4=\left(a-4\right)\left(2a-1\right)-4\left(2a-1\right)\)
\(=\left(a-8\right)\left(2a-1\right)\)
a) `4abc-8ab^2c=4abc(1-2b)`
b) `x^2 (2a-1)+x(1-2a) = x^2 (2a-1) -x(2a-1) = (2a-1)(x^2-x)=x(2a-1)(x-1)`
c) `9a^4 (a-2) +a^2 (a-2) = (a-2)(9a^4+a^2)=a^2 (a-2)(9a^2+1)`
d) `(a-4)(2a-1)-8a+4=(a-4)(2a-1)-4(2a-1)=(2a-1)(a-8)`
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.
$3x(x-2)-x+2=3x(x-2)-(x-2)=(x-2)(3x-1)$
\(a,ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\\ =a^2b-ab^2+b^2c-bc^2+ca\left(c-a\right)\\ =\left(a^2b-bc^2\right)-\left(ab^2-b^2c\right)+ca\left(c-a\right)\\ =b\left(a-c\right)\left(a+c\right)-b^2\left(a-c\right)-ca\left(a-c\right)\\ =\left(a-c\right)\left(ab+bc-b^2-ca\right)\\ =\left(a-c\right)\left(b-c\right)\left(a-b\right)\)
\(b,x^2-3xy-10y^2\\ =x^2+2xy-5xy-10y^2\\ =x\left(x+2y\right)-5y\left(x+2y\right)=\left(x-5y\right)\left(x+2y\right)\)
\(c,3x\left(x-2\right)-x+2=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)
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: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
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) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
TL :
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