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a,x2-y2-2x+2y
= (x+y)(x-y) - 2(x-y)
= (x-y)(x+y-2)
b,2x+2y-x2-xy
= 2(x+y) - x(x+y)
= (x+y)(2-x)
c,3a2-6ab+3b2-12c2
= 3(a2 - 2ab + b2 - 4c2)
= 3[(a-b)2 - 4c2)
= 3(a-b-2c)(a-b+2c)
d,x2-25+y2+2xy
= (x+y)2 - 25
= (x+y+5)(x+y-5)
e) a2+2ab+b2-ac-bc
= (a+b)2-c(a+b)
= (a+b)( a+b-c)
f) x2-2x-4x2-4y
= -3x2-2x-4y
= -(3x2+2x+4y)
g)x2y-x3-9y+9x
= x2(y-x)-9(y-x)
= (y-x)(x2-9)
h) x2(x-1)+16(1-x)
= x2(x-1)-16(x-1)
= (x-1)(x2-16)
= (x-1)(x-4)(x+4)
n) 81x2-6yz-9y2-z2
= (9x)2-[(3y)2+6yz+z2]
=(9x)2-(3y+z)2
=(9x+3y+z)(9x-3y-z)
m) xz- yz-x2+2xy-y2
= z(x-y)-(x2-2xy+y2)
= z(x-y)-(x-y)2
= (x-y)(z-x+y)
p) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x+3) + 5(x+3)
= (x+3)(x+5)
k) x2 - x - 12
= x2 + 3x - 4x - 12
= x(x+3) - 4(x+3)
= (x+3)(x-4)
a: \(3abc^3-6a^2b^3c+12a^3bc\)
\(=3abc\cdot c^2-3abc\cdot2ab^2+3abc\cdot4a^2\)
\(=3abc\left(c^2-2ab^2+4a^2\right)\)
b: \(27-8y^3\)
\(=3^3-\left(2y\right)^3\)
\(=\left(3-2y\right)\left(9+6y+4y^2\right)\)
c: Sửa đề: \(4x^2+4x-y^2+1\)
\(=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+1+y\right)\left(2x+1-y\right)\)
d: \(3a^2\cdot\left(x-2\right)-6ab\cdot\left(2-x\right)\)
\(=3a^2\cdot\left(x-2\right)+6ab\cdot\left(x-2\right)\)
\(=\left(x-2\right)\left(3a^2+6ab\right)\)
\(=3a\left(a+2b\right)\left(x-2\right)\)
\(5a^2-14ab-3b^2\\ =5a^2-15ab-ab-3b^2\\ =5a\left(a-3b\right)-b\left(a-3b\right)\\ =\left(5a-b\right)\left(a-3b\right)\)
\(5a^2-14ab-3b^2\)
\(=5a^2-15ab+ab-3b^2\)
\(=5a\left(a-3b\right)+b\left(a-3b\right)\)
\(=\left(a-3b\right)\left(5a+b\right)\)
\(9a^2b+6ab^2+b^3-6ab-2b^2\)
\(=b\left(9a^2+6ab+b^2-6a-2b\right)\)
\(=b\left[\left(3a+b\right)^2-2\left(3a+b\right)\right]\)
\(=b\left(3a+b\right)\left(3a+b-2\right)\)
\(=b\left(9a^2+6ab+b^2\right)-2b\left(3a+b\right)\)
\(=b\left(3a+b\right)^2-2b\left(3a+b\right)\)
\(=b\left(3a+b\right)\left(3a+b-2\right)\)
\(9a^2b+6ab^2+b^3-6ab-2b^2\)
\(=b\left(9a^2+6ab+b^2-6a-2b\right)\)
\(=b\left[\left(3a+b\right)^2-2\left(3a+b\right)\right]\)
\(=b\left(3a+b\right)\left(3a+b-2\right)\)
b: \(2x^2-7xy+3y^2+x-3y\)
\(=2x^2-6xy-xy+3y^2+x-3y\)
\(=2x\left(x-3y\right)-y\left(x-3y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y+1\right)\)
Lời giải:
a.
Đặt $2a^2+5ab-3b^2-7b-2=(a+mb+n)(2a+pb+k)$ với $m,n,p,k$ nguyên
$\Leftrightarrow 2a^2+5ab-3b^2-7b-2=2a^2+ab(2m+p)+mpb^2+a(k+2n)+b(km+np)+kn$
Đồng nhất hệ số:
\(\left\{\begin{matrix} 2m+p=5\\ mp=-3\\ k+2n=0\\ km+np=-7\\ kn=-2\end{matrix}\right.\)
Giải hpt này ta thu được $m=3; n=1; p=-1; k=-2$
Vậy $2a^2+5ab-3b^2-7b-2=(a+3b+1)(2a-b-2)$
b. Đa thức không phân tích được thành nhân tử
b: Ta có: \(2x^2-7xy+3y^2+x-3y\)
\(=2x^2-6xy-xy+3y^2+x-3y\)
\(=2x\left(x-3y\right)-y\left(x-3y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y+1\right)\)
a^3+b^3+6ab-8=(a + b - 2) (a² - a b + 2a + b² + 2b + 4)
3a2 - 6ab + 3b2 - 12c2
= 3a2 - 3a x 2b + 3b2 - 3 x 4c2
= 3(a2 -a x b +b2 - 4c2)
phân tích thành nhân tử :
3a2 - 6ab + 3b2 - 12c2
=3(a\(^2\)-2ab+b\(^2\)-4c\(^2\))
=3[(a-b)\(^2\)-(2c)\(^2\)]
=3(a-b-2c)(a-b+2c)