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Em tham khảo:
cho 3 số x,y,z đôi một khác nhau và x+y+z=0 Tính\(P=\dfrac{2018\left(x-y\right)\left(y-z\right)\left(z-x\right)}{2xy^2+2... - Hoc24
Ta có \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Leftrightarrow\left(x+y\right)^3=-z^3\)
\(\Leftrightarrow x^3+y^3+z^3+3xy\left(x+y\right)=0\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)
Đặt \(A=2xy^2+2yz^2+2zx^2+3xyz=2xy^2+2yz^2+2zx^2+x^3+y^3+z^3\)
\(=x^2\left(2z+x\right)+y^2\left(2x+y\right)+z^2\left(2y+z\right)\)
Do \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}2z+x=z-y\\2x+y=x-z\\2y+z=y-x\end{matrix}\right.\)
\(\)\(\Rightarrow A=x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\)
\(=x^2\left(z-y\right)-y^2\left(z-y+y-x\right)+z^2\left(y-x\right)\)
\(=\left(x^2-y^2\right)\left(z-y\right)-\left(z^2-y^2\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(z-y\right)\left(x+y-z-y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
\(\Rightarrow\dfrac{2018\left(x-y\right)\left(y-z\right)\left(x-z\right)}{A}=2018\)
\(\Rightarrow P=2018\)
Vậy \(P=2018\)
Chắc đề là \(x+y+z=3\)
Ta có:
\(\left(2x+y+z\right)^2=\left(x+y+x+z\right)^2\ge4\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow P\le\dfrac{x}{4\left(x+y\right)\left(x+z\right)}+\dfrac{y}{4\left(x+y\right)\left(y+z\right)}+\dfrac{z}{4\left(x+z\right)\left(y+z\right)}\)
\(\Rightarrow P\le\dfrac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{4\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\dfrac{xy+yz+zx}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Mặt khác:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(zy+yz+zx\right)=\dfrac{8}{3}\left(xy+yz+zx\right)\)
\(\Rightarrow P\le\dfrac{xy+yz+zx}{2.\dfrac{8}{3}\left(xy+yz+zx\right)}=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Lời giải:
$xy+yz+xz=\frac{1}{2}[(x+y+z)^2-(x^2+y^2+z^2)]=\frac{1}{2}(a^2-b^2)$
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}$
$\Rightarrow xyz=c(xy+yz+xz)=\frac{1}{2}c(a^2-b^2)$
Khi đó:
$P=(x+y+z)^3-3(x+y)(y+z)(x+z)$
$=(x+y+z)^3-3[(x+y+z)(xy+yz+xz)-xyz]=(x+y+z)^3-3(xy+yz+xz)(x+y+z)+3xyz$
$=a^3-\frac{3}{2}a(a^2-b^2)+\frac{3}{2}c(a^2-b^2)$
