Câu hỏi của Mai Huy Bảo

Question

phân tích đa thức thành nhân tử  ( x + y - z)3 - x3 - y3 + z3

Toru · Accepted Answer

$\left(x+y-z\right)^3-x^3-y^3+z^3$$=\left[\left(x+y\right)-z\right]^3-x^3-y^3+z^3$$=\left(x+y\right)^3-z^3-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3$$=x^3+y^3-z^3+3xy\left(x+y\right)-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3$$=3xy\left(x+y\right)-3z\left(x+y\right)\left(x+y-z\right)$$=3\left(x+y\right)\left[xy-z\left(x+y-z\right)\right]$$=3\left(x+y\right)\left(xy-zx-yz+z^2\right)$$=3\left(x+y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]$$=3\left(x+y\right)\left(y-z\right)\left(x-z\right)$#$Urushi\text{☕}$Câu trả lời: $\left(x+y-z\right)^3-x^3-y^3+z^3$$=\left[\left(x+y\right)-z\right]^3-x^3-y^3+z^3$$=\left(x+y\right)^3-z^3-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3$$=x^3+y^3-z^3+3xy\left(x+y\right)-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3$$=3xy\left(x+y\right)-3z\left(x+y\right)\left(x+y-z\right)$$=3\left(x+y\right)\left[xy-z\left(x+y-z\right)\right]$$=3\left(x+y\right)\left(xy-zx-yz+z^2\right)$$=3\left(x+y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]$$=3\left(x+y\right)\left(y-z\right)\left(x-z\right)$#$Urushi\text{☕}$

Minh Duong · Answer

Áp dụng (a+b)3 = a3+b3+3ab(a+b), ta có:(x+y+z)3-x3-y3-z3=[(x+y)+z]3-x3-y3-z3=(x+y)3+z3+3z(x+y)(x+y+z)-x3-y3-z3=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)-x3-y3-z3=3(x+y)(xy+xz+yz+z2)=3(x+y)[x(y+z)+z(y+z)]=3(x+y)(y+z)(x+z)Câu trả lời: Áp dụng (a+b)3 = a3+b3+3ab(a+b), ta có:(x+y+z)3-x3-y3-z3=[(x+y)+z]3-x3-y3-z3=(x+y)3+z3+3z(x+y)(x+y+z)-x3-y3-z3=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)-x3-y3-z3=3(x+y)(xy+xz+yz+z2)=3(x+y)[x(y+z)+z(y+z)]=3(x+y)(y+z)(x+z)

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