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a: =(a^2-b^2)-(2a-2b)
=(a-b)(a+b)-2(a-b)
=(a-b)(a+b-2)
b: =(3x-3y)+5y(x-y)
=3(x-y)+5y(x-y)
=(x-y)(5y+3)
c: \(=\left(x+y\right)^2\left(x-y\right)+x\left(y-x\right)\)
=(x-y)*(x+y)^2-x(x-y)
=(x-y)[(x+y)^2-x]
d: \(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)
=(-x-4y+5)(3x+2y+3)
e: =16-(x^2-4xy+4y^2)
=16-(x-2y)^2
=(4-x+2y)(4+x-2y)
g: =9x^2-6x+1-(3xy-y)
=(3x-1)^2-y(3x-1)
=(3x-1)(3x-y-1)
h: =(x-y)^3-z^3
=(x-y-z)[(x-y)^2+z(x-y)+z^2]
=(x-y-z)(x^2-2xy+y^2+xz-yz+z^2)
a) \(a^2-b^2-2a+2b\)
\(=\left(a^2-b^2\right)-\left(2a-2b\right)\)
\(=\left(a+b\right)\left(a-b\right)-2\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-2\right)\)
b) \(3x-3y-5x\left(y-x\right)\)
\(=\left(3x-3y\right)+5x\left(x-y\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
c) \(x\left(x+y\right)^2-y\left(x+y\right)^2+xy-x^2\)
\(=\left(x+y\right)^2\left(x-y\right)+\left(xy-x^2\right)\)
\(=\left(x+y\right)^2\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+2xy+y^2-x\right)\)
d) \(\left(x-y+4\right)^2-\left(2x+3y-1\right)\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
a, (\(x\) + y).(\(x\) + y)2 - 3\(xy\).(\(x\) + y)
= (\(x+y\))3 - 3\(x^2\)y - 3\(xy^2\)
= \(x^3\) + 3\(x^2\).y + 3\(xy^2\) + y3 - 3\(x^2\).y - 3\(xy^2\)
= \(x^3\) + y3
b, (\(x-y\)).(\(x-y\))2 - 3\(xy\).(\(x-y\))
= (\(x\) - y)3 - 3\(x^2\).y + 3\(xy^2\)
= \(x^3\) - 3\(x^2\)y + 3\(xy^2\) - y3 - 3\(x^2\)y + 3\(xy^2\)
= \(x^3\) - 6\(x^2\)y + 6\(xy^2\) - y3
Bài làm :
\(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
\(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
\(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
\(d ) x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
a) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
b) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
c) \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
a: M=2(-2x-3xy^2+1)-3xy^2+1
=-4x-6xy^2+2-3xy^2+1
=-4x-9xy^2+3
b: Thay x=-2 và y=3 vào M, ta được:
M=2*(-2)-3*(-2)*3^2+1
=-4+1+6*9
=54-3
=51
a) Nhóm x^2 và y^2 ; x và y
b) Nhóm 3 hạng tử đầu lại vs nhau . Sau cùng xuất hiện nhân tử chung là 3
c) Nhóm 2 hạng tử đầu với nhau. ba hạng tử còn lại với nhau .
d) .....
D,ghép đầu với cuối là hằng dẳng thức 2 cái giữa với nhau là nhân tử chung là 3x