Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2+2xy+x+2y\)
\(=x\left(x+1\right)+2y\left(x+1\right)\)
\(=\left(x+1\right)\left(2y+x\right)\)
\(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
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) \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(1,a^2-2a+1-b^2\)
\(=\left(a^2-2a+1\right)-b^2\)
\(=\left(a-1\right)^2-b^2\)
\(=\left(a-1-b\right)\left(a-1+b\right)\) Khai triển thành hằng đẳng thức số 3 e nhé.
\(2,x^2+2xy+y^2-81\)
\(=\left(x^2+2xy+y^2\right)-81\)
\(=\left(x+y\right)^2-9^2\)
\(=\left(x+y-9\right)\left(x+y+9\right)\)Cái này cũng HĐT số 3 nè
\(3,x^2+6y-9-y^2\)
\(=-\left(y^2-6y+9\right)+x^2\)
\(=-\left(y-3\right)^2+x^2\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
\(5,4x^2+y^2-9-4xy\)
\(=\left(4x^2-4xy+y^2\right)-9\)
\(=\left(2x-y\right)^2-3^2\)
\(=\left(2x-y-3\right)\left(2x-y+3\right)\)
Học tốt
\(a,49.\left(y-4\right)^2-9y^2-36y-36=49\left(y-4\right)^2-9\left(y^2+4y+4\right)\)
\(=49\left(y-4\right)^2-9\left(y+4\right)^2=\left(7y-28\right)^2-\left(3y+12\right)^2\)
\(=\left(7y-28+3y+12\right)\left(7y-28-3y-12\right)\)
\(=\left(10y-16\right)\left(4y-40\right)=8\left(5y-8\right)\left(y-10\right)\)
\(b,xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=xyz-xy-yz-xz+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-x-y+1\right)\)
\(=\left(z-1\right)\text{[}x\left(y-1\right)-\left(y-1\right)\text{]}\)
\(=\left(z-1\right)\left(y-1\right)\left(x-1\right)\)
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) \(49-x^2+2xy-y^2=7^2-\left(x-y\right)^2=\left(7+x-y\right)\left(7-x+y\right)\)
\(x^2+2xy+7x+7y+y^{2+10}\)
\(\text{phân tích đa thức thành nhân tử}\)
\(y^{12}+2xy+7y+x^2+7x\)
tách \(^{x^2}\)ra rồi làm thừa số chung, toán SGK đem ra hỏi làm j
\(B=\left(2xy-x^2-y^2+z^2\right)\left(2xy+x^2+y^2-z^2\right)\)
\(=\left[z^2-\left(x-y\right)^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(z-x+y\right)\left(z+x-y\right)\left(x+y-z\right)\left(x+y+z\right)\)
a) \(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
x2−x−y2−y=(x2−y2)−(x+y)=(x−y)(x+y)−(x+y)=(x+y)(x−y−1)