Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(x^3+x^2z-xyz+y^2z+y^3\)
\(=\left(x^3+y^3\right)+\left(x^2z-xyz+y^2z\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)\)
\(=\left(x+y+z\right)\left(x^2-xy+y^2\right)\)
\(=0\cdot\left(x^2-xy+y^2\right)\)
\(=0\left(dpcm\right)\)
a) \(B=x^3+x^2z+y^2z-xyz+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)\)
\(=\left(x^2-xy+y^2\right)\left(x+y+z\right)\)
b) \(B=\left(x^2-xy+y^2\right)\left(x+y+z\right)=x^2-xy+y^2\)
\(=x^2-2.x.\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\)
Dấu bằng xảy ra khi \(x=y=0\)
k: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
i: \(=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)
\(g,27+27x+9x^2+x^3=\left(3+x\right)^3\\ i,2x^2+2y^2-x^2z+z-y^2z-2=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)=\left(x^2+y^2-1\right)\left(2-z\right)\)
\(k,8-27x^2=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(l,3x^2-6xy+3y^2=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)
\(x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
\(\Leftrightarrow x^2+y^2+z^2-xy-xz-yz=0\)
\(\Leftrightarrow x=y=z\)
A = \(\left(x^3+y^3\right)+\left(x^2z+y^2z-xyz\right)=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)=\left(x^2-xy+y^2\right)\left(x+y+z\right)=\left(x^2-xy+y^2\right).0=0\)Kuroba Kaito = Kaito Kid :D
thanks