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a)x^2-4xy+4y^2-4
=(x2-4xy+4y2)-4
=(x-2y)2-4
=(x-2y+2)(x-2y-2)
b)16-x^2+2xy-y^2
=16-(x2-2xy+y2)
=16-(x-y)2
=[4-(x-y)][4+(x-y)]
=(4-x+y)(4+x-y)
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
a 4x -4y +(x-y)^2
=4(x-y)+(x-y).(x-y)
=(x-y).(4+x-y)
c x^2(x+1)-4(x+1)
(x+1).(x^2-4)
d x^4-(x^2-2x+1)
=x^4-(x-1)^2
=x^2(x-x+1)(x-x-1)
MIK KO BIT DUNG HAY KO CON B THI MIK KO BIET LAM
Câu b dễ thôi
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
2x(x-2)+2y(x-2)= (x-2)(2x+2y)=2(x-2)(x+y)
b,2(xy+xyz-2x-2z)
c, 3(x^2-xy-x-y)
a) Ta có : 2x2 - 4x + 2xy - 4y
= 2x(x - 2) + 2y(x - 2)
= (x - 2)(2x + 2y)
= 2(x - 2)(x + y)
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
a)x^2+2x-4y^2-4y
=(x2-4y2)+(2x-4y)
=(x-2y)(x+2y)+2.(x-2y)
=(x-2y)(x+2y+2)
b)x^4-6x^3+54x-81
=(x4-81)+(-6x3+54x)
=(x2-9)(x2+9)-6x.(x2-9)
=(x2-9)(x2+9-6x)
=(x-3)(x+3)(x-3)2
=(x-3)3(x+3)
c)ax^2+ax-bx^2-bx-a+b
=(ax2-bx2)+(ax-bx)+(-a+b)
=x2.(a-b)+x.(a-b)-(a-b)
=(a-b)(x2+x+1)
\(a,2x+4y=2\left(x+2y\right)\)
\(b,x^2+2xy+y^2-1\)
\(=\left(x+y\right)^2-1\)
\(=\left(x+y-1\right)\left(x+y+1\right)\)