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a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
a) x^2 - 5xy +4y^2= x^2 -xy -4xy+4y^2= (x^2-xy) - (4xy - 4y^2)= x(x-y)-4y(x-y)=(x-y)*(x - 4y)
b) x^2 -y^4+9y -x(9+y-y^3= x^2-y^4 +9y-9x-xy+xy^3= (x^2-xy)-(9x-9y)+(xy^3-y^4)=x(x-y)-9(x-y)+y^3(x-y)=(x-y)*(y^3+x-9)
d) 2u^2+2v^2-5uv=(2u^2-4uv)+(2v^2-uv)=2u(u-2v)+v(2v-u)= 2u(u-2v)-v(u-2v)=(u-2v)*(2u-v)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
1) \(x^2+2xy+y^2-x-y-12\)
= \(\left(x+y\right)^2-\left(x+y\right)-12\)
Đặt \(x+y=z\) (đặt ẩn phụ)
\(\Rightarrow z^2-z-12\)
\(=z^2+3z-4z-12\)
\(=z\left(z+3\right)-4\left(z+3\right)\)
\(=\left(z+3\right)\left(z-4\right)\)
Khi đó: \(\left(x+y+3\right)\left(x+y-4\right)\)
#HuyenAnh
1) Ta có: \(x^2-2x-9y^2+6y\)
\(=x^2-2x+1-9y^2+6y-1\)
\(=\left(x-1\right)^2-\left(3y-1\right)^2\)
\(=\left(x-1-3y+1\right)\left(x-1+3y-1\right)\)
\(=\left(x-3y\right)\left(x+3y-2\right)\)
3) Ta có: \(x^2-9-4xy+4y^2\)
\(=\left(x-2y\right)^2-3^2\)
\(=\left(x-2y-3\right)\left(x-2y+3\right)\)
4) Ta có: \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=2b\cdot2a=4ab\)
5) Ta có: \(\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-3xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)
6) Ta có: \(\left(x-y\right)^3+3xy\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)^2+3xy\right]\)
\(=\left(x-y\right)\left(x^2-2xy+y^2+3xy\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\)
7) Ta có: \(4x^2-12x-46\)
\(=\left(2x\right)^2-2\cdot2x\cdot3+9-55\)
\(=\left(2x-3\right)^2-55\)
\(=\left(2x-3-\sqrt{55}\right)\left(2x-3+\sqrt{55}\right)\)