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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
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)
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: \(x^2-2xy+y^2+4x-4y-5=\left(x^2+y^2+2^2-2xy+4x-4y\right)-9\)
\(=\left(y-x-2\right)^2-3^2=\left(y-x-2-3\right)\left(y-x-2+3\right)=\left(y-x-5\right)\left(y-x+1\right)\)
s) = ( x2 - 2xy + y2 ) - ( 2xy )2 = ( x - y - 2xy )( x - y + 2xy )
u) sửa +4y thành -4y
= 4( x - y ) - x2( x - y ) = ( x - y )( 2 - x )( 2 + x )
\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)
\(=\left(x-y+2\right)^2-9\)
\(=\left(x-y+2\right)^2-3^2\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
nhớ nha
\(x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)
\(=\left(x-y+2\right)^2-9\)
\(=\left(x-y+2-3\right)\left(x-y+2+3\right)\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
\(x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2-1+4\left(x-y-1\right)\)
\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1+4\right)\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
mk làm lun nha
(x-y)^2-3^2
=(x-y-3)(x-y+3)
các câu còn lại tương tự
**** cho mk nha
a)x2-2xy+y2-9
=(x-y)2-32
=(x-y-3)(x-y+3)
b)2x3+4x2y+2xy2
=2x(x2+2xy+y2)
=2x(x+y)2
c)x2-4xy+4y2-36z2
=(x-2y)2-(6z)2
=(x-2y-6z)(x-2y+6z)
a) x2 - y2 - 4x + 4y
= (x2 - 4x + 4) - (y2 - 4y + 4)
= (x - 2)2 - (y - 2)2
= (x - 2 - y + 2)(x - 2 + y - 2)
= (x - y)(x + y - 4)
b) (xy + 4)2 - 4(x + y)2
= (xy + 4)2 - [2(x + y)]2
= (xy + 4)2 - (2x + 2y)2
= (xy + 4 - 2x - 2y)(xy + 4 + 2x + 2y)
c) 25 - x2 + 2xy - y2
= 25 - (x2 - 2xy + y2)
= 52 - (x - y)2
=> (5 - x + y)(5 + x - y)
a) \(x^2-y^2-4x+4y=\left(x^2-y^2\right)-\left(4x-4y\right)=\left(x+y\right)\left(x-y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-4\right)\)
b) \(\left(xy+4\right)^2-4\left(x+y\right)^2=\left(xy+4\right)^2-\left(2x+2y\right)^2=\left(xy+4+2x+2y\right)\left(xy+4-2x-2y\right)\)
c) \(25-x^2+2xy-y^2=25-\left(x^2-2xy+y^2\right)=5^2-\left(x-y\right)^2=\left(5+x-y\right)\left(5-x+y\right)\)
x2+4x-2xy-4y+y2
=(x2-2xy+y2)+(4x-4y)
=(x-y)2+4(x-y)
=(x-y)(x-y+4)