ta có: (x+y+z)2=x2+y2+z2+2(xy+yz+xz)
do đó ta đặt: x2+y2+z2=a
xy+yz+xz=b
=> (x2+y2+z2)*(x+y+x)2+(xy+yz+xz)2
=a*(a+2b)+b2
=(a+b)2
thay a=x^2+y^2+z^2,b=xy+yz+xz vào
ta được (x^2+y^2+z^2+xy+yz+xz)^2
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
Cho 1/x + 1/y + 1/z = 0. Tính N = yz/x2 + zx/y2 + xy/z2
Cho x^2+y^2+z^2=10. Tính P= (xy=yz=xz)^2 + (x^2-yz)^2 + (y^2 -xz)^2 + (z^2-xy)^2
\(x^2+y^2+z^2-xy-yz-xz\)
\(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
Phân tích đa thức thành nhân tử
xy(x+y) + yz(y+z) +xz(x+z) + 2xyz
x2+5x+6
7x-6x2-2
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
Phân tích đa thức thành nhân tử:
xy(x+y) - yz(z+y)+xz(x-z)
