Phân tích đa thức thành nhân tửX3+y3+z3-3xyz(X2+xy)2

\(\left(x^2+xy\right)^2-\left(y^2+xy\right)^2\)\(=\left(x^2+xy-y^2-xy\right)\left(x^2+xy+y^2+xy\right)\)\(=\left(x^2-y^2\right)\left(x^2+2xy+y^2\right)\)\(=\left(x-y\right)\left(x+y\right)\left(x+y\right)^2\)\(=\left(x-y\right)\left(x+y\right)^3\)Câu trả lời: \(\left(x^2+xy\right)^2-\left(y^2+xy\right)^2\)\(=\left(x^2+xy-y^2-xy\right)\left(x^2+xy+y^2+xy\right)\)\(=\left(x^2-y^2\right)\left(x^2+2xy+y^2\right)\)\(=\left(x-y\right)\left(x+y\right)\left(x+y\right)^2\)\(=\left(x-y\right)\left(x+y\right)^3\)

•x3+y3+z3-3xyz=(x+y)3-3xy(x+y)+z3-3xyz=(x+y+z)[(x+y)2-(x+y).z+z2]-3xy(x+y+z)=(x+y+z)(x2+y2+z2+2xy-xz-yz) -3xy(x+y+z)=(x+y+z)(x2+y2+z2-xy-yz-xz)•(x2+xy)2-(y2+xy)2=[x(x+y)]2-[y(x+y)]2=x2.(x+y)2-y2.(x+y)2=(x+y)2.(x2-y2)=(x+y)2.(x+y).(x-y)=(x+y)3(x-y)•3x2-3x-36=3.(x2-x-12)=3(x2-4x+3x-12)=3[x(x-4)+3(x-4)]=3(x-4)(x+3)Câu trả lời: •x3+y3+z3-3xyz=(x+y)3-3xy(x+y)+z3-3xyz=(x+y+z)[(x+y)2-(x+y).z+z2]-3xy(x+y+z)=(x+y+z)(x2+y2+z2+2xy-xz-yz) -3xy(x+y+z)=(x+y+z)(x2+y2+z2-xy-yz-xz)•(x2+xy)2-(y2+xy)2=[x(x+y)]2-[y(x+y)]2=x2.(x+y)2-y2.(x+y)2=(x+y)2.(x2-y2)=(x+y)2.(x+y).(x-y)=(x+y)3(x-y)•3x2-3x-36=3.(x2-x-12)=3(x2-4x+3x-12)=3[x(x-4)+3(x-4)]=3(x-4)(x+3)