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Giải
\(x^3+y^3-z^3+3xyz\)
= \(\left(x+y\right)^3-z^3-3x^2y-3xy^2+3xyz\)
= \(\left(x+y-z\right)\left[\left(x +y\right)^2+\left(x+y\right)z+z^2\right]-3xy\left(x+y-z\right)\)
= \(\left(x+y-z\right)\left(x^2+y^2+z^2-xy+xz+yz\right)\)
Vậy \(x^3+y^3-z^3+3xyz\) chia hết cho x + y - z và được thương là:
\(x^2+y^2+z^2-xy+xz+yz\)
\(\dfrac{x^3+y^3-z^3+3xyz}{x+y-z}\)
\(=\dfrac{\left(x+y\right)^3-z^3-3xy\left(x+y\right)+3xyz}{x+y-z}\)
\(=\dfrac{\left(x+y-z\right)\left(x^2+2xy+y^2+xz+yz+z^2\right)-3xy\left(x+y-z\right)}{x+y-z}\)
\(=x^2+y^2+z^2-xy+xz+yz\)
Bài 1:
Ta có:\(x^2+xy+y^2+1\)
\(=x^2+\dfrac{1}{2}xy+\dfrac{1}{2}xy+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+1\)
\(=\left(x^2+\dfrac{1}{2}xy\right)+\left(\dfrac{1}{2}xy+\dfrac{1}{4}y^2\right)+\dfrac{3}{4}y^2+1\)
\(=x.\left(x+\dfrac{1}{2}y\right)+\dfrac{1}{2}y.\left(x+\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2+1\)
\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x+\dfrac{1}{2}y\right)^2\ge0;\dfrac{3}{4}y^2\ge0\)
\(\Rightarrow\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\Rightarrow\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1\ge1>0\)
Hay \(x^2+xy+y^2+1>0\) (đpcm)
Chúc bạn học tốt!!!
\(x^3+y^3-z^3+3xyz\)
\(=\left(x+y\right)^3-z^3-3xy\left(x+y\right)+3xyz\)
\(=\left(x+y-z\right)\left(x^2+2xy+y^2+xz+yz+z^2\right)-3xy\left(x+y-z\right)\)
\(=\left(x+y-z\right)\left(x^2+y^2+z^2-xy+xz+yz\right)⋮x+y-z\)
Ta có: x+y+z=0⇔x+y=−z
⇔(x+y)3=(−z)3
⇔x3+3x2y+3xy2+y3=−z3
⇔x3+y3+z3=−3x2y−3xy2
⇔x3+y3+z3=−3xy(x+y)
⇔x3+y3+z3=−3xy(−z)=3xyz(đpcm)
Ta có : x+y+z = 0
\(\Rightarrow x+y=-z\)
\(\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=\left(-z\right)^3\)
\(\Leftrightarrow x^3+y^3+z^3=-3x^2y-3xy^2\)
\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(x+y\right)\)
\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(-z\right)\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)
Lời giải :
\(x+y+z=0\)
\(\Leftrightarrow x+y=-z\)
\(\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=-z^3\)
\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(x+y\right)\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)(đpcm)