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a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
Xin lỗi, mjk làm lại nhé *(^o^)*
x2 - 2xy + y2 + 4x - 4y - 5
= (x - y)2 + 4(x - y) - 5
Đặt x - y = t, ta có:
t2 + 4t - 5
= t2 - t + 5t - 5
= t(t - 1) + 5(t - 1)
= (t - 1)(t + 5)
= (x - y - 1)(x - y + 5)
x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử
= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung
= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung
Cách 1: \(x^2-2xy+y^2+4x-4y-5=\left(y^2-xy+y\right)+\left(-xy+x^2-x\right)+\left(-5y+5x-5\right)\)
\(=y\left(y-x+1\right)-x\left(y-x+1\right)-5\left(y-x+1\right)=\left(y-x+1\right)\left(y-x-5\right)\)
Cách 2: \(x^2-2xy+y^2+4x-4y-5=\left(x^2+y^2+2^2-2xy+4x-4y\right)-9\)
\(=\left(y-x-2\right)^2-3^2=\left(y-x-2-3\right)\left(y-x-2+3\right)=\left(y-x-5\right)\left(y-x+1\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)
\(=\left(x-y+2\right)^2-9\)
\(=\left(x-y+2\right)^2-3^2\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
nhớ nha
\(x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)
\(=\left(x-y+2\right)^2-9\)
\(=\left(x-y+2-3\right)\left(x-y+2+3\right)\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
\(x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2-1+4\left(x-y-1\right)\)
\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1+4\right)\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
x2-2xy+y2+4x-4y-5 = (x2-2xy+y2)+(4x-4y)-5
= (x+y)2+4(x-y)-5
= 4(x-y)-5
Làm đại, không chắc lắm!
x2+4x-2xy-4y+y2
=(x2-2xy+y2)+(4x-4y)
=(x-y)2+4(x-y)
=(x-y)(x-y+4)
=( x2-2xy +y2)+(4x-4y)=(x-y)2+4(x-y)=(x-y)(x-y+4)
(x-y)^2+4(x-y)+4-9
(x-y+2)^2-9
(x-y-1)(x-y+5)