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a) \(x^3+3x^2+3x+1=\left(x+1\right)^3\)
b) \(x^3-6x^2+12x-8=\left(x-2\right)^3\)
c) \(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
d) \(49-x^2+2xy-y^2=7^2-\left(x-y\right)^2=\left(7+x-y\right)\left(7-x+y\right)\)
\(x^2-y^2\)
\(=x^2+xy-xy-y^2\)
\(=x\left(x+y\right)-y\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)\)
\(x^2+2xy+x+2y\)
\(=x\left(x+1\right)+2y\left(x+1\right)\)
\(=\left(x+1\right)\left(2y+x\right)\)
\(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
a)x2+2xy+x+2y
=(2xy+x2)+(2y+x)
=x(2y+x)+(2y+x)
=(x+1)(2y+x)
b)7x2-7xy-5x+5y
=(5y-7xy)+(7x2-5x)
=y(5-7x)-x(5-7x)
=(5-7x)(y-x)
c)x2-6x+9-9y2
=(x2+3xy-3x)-(3xy+9y2-9y)-(3x+9y-9)
=x(x+3y-3)-3y(x+3y-3)-3(x+3y-3)
=(x-3y-3)(x+3y-3)
d)x3-3x2+3x-1+2(x2-x)
Ta thấy x=1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x-1
=(x-1)(x2+1)
e) (x+y)(y+z)(z+x)+xyz
đề sai
f)x(y2-z2)+y(z2-x2)
=(xy2+yz2)+(x2y+xz2)
=y(xy+z2)-x(xy+z2)
=(y-x)(xy+z2)
a) \(a^3+a^2b-a^2c-abc=a^2\left(a+b\right)-ac\left(a+b\right)=a\left(a+b\right)\left(a-c\right)\)
b) mk chỉnh lại đề
\(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
c) \(4-x^2-2xy-y^2=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\)
d) \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
x3-x2+x+3=x3+1-x2+1+x+1
=(x+1)(x2+x+1)-(x2-1)+(x+1)
=(x+1)(x2+x+1)-(x+1)(x-1)+(x+1)
=(x+1)[(x2+x+1)-(x-1)+1]
=(x+1)(x2+x+1-x+1+1)
=(x+1)(x2+3)
a) A+\(\left(x^2-4xy^2+2xz-3y^2\right)\)=0
=> A=0-\(\left(x^2-4xy^2+2xz-3y^2\right)\)
=>A=\(-x^2+4xy^2-2xz+3xy^2\)
b)B+\(\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
=>B=\(6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
=>B=\(6x^2+9xy-y^2-5x^2+2xy\)
=>B=\(\left(6x^2-5x^2\right)+\left(9xy+2xy\right)-y^2\)
=>B=\(x^2+11xy-y^2\)
c)ta có:
B+(\(4x^2y+5y^2-3xz+z^2\))
thay B=\(x^2+11xy-y^2\) vào biểu thức trên ta được:
\(x^2+11xy-y^2\) + (\(4x^2y+5y^2-3xz+z^2\))
= \(x^2+11xy-y^2+4x^2y+5y^2-3xz+z^2\)
=\(\left(5y^2-y^2\right)+x^2+11xy-y^2+4x^2y+5y^2-3xz+z^2\)
=\(4y^2+x^2+11xy-y^2+4x^2y+5y^2\)
đúng chưa bạn
a) A + (x\(^2\) - 4xy\(^2\) + 2xz - 3y\(^2\) ) = 0
(=) A = -x\(^2\) +4xy\(^2\) - 2xz + 3y\(^2\)
b) B + (5x\(^2\) - 2xy) = 6x\(^2\) + 9xy - y\(^2\)
(=) B = 6x\(^2\) + 9xy - y\(^2\) - 5x\(^2\) + 2xy
(=) B = x\(^2\) + 11xy - y\(^2\)
c) Đa thức không chứa biến x là 5 ( có thể thay đổi )
(=) B + (4x\(^2\)y + 5y\(^2\) - 3xz + z\(^2\)) = 5
(=) B = 5 - 4x\(^2\)y - 5y\(^2\) + 3xz - z\(^2\)
1. A+(x2-4xy2+2xz-3y2) =0
=> A = -(x2-4xy2+2xz-3y2)
=> A = -x2+4xy2-2xz+3y2
Vậy A=-x2+4xy2-2xz+3y2
\(x^2-3y^2-8z^2+2xy-10yz+2xz\)
\(=x^2-3y^2-8z^2+3xy-xy-4yz-6yz+4xz-2xz\)
\(=\left(x^2+3xy+4xz\right)+\left(-xy-3y^2-4yz\right)+\left(-2xz-6yz-8z^2\right)\)
\(=x\left(x+3y+4z\right)-y\left(x+3y+4z\right)-2z\left(x+3y+4z\right)\)
\(=\left(x+3y+4z\right)\left(x-y-2z\right)\)