\(D=\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{zx+z+1}\)

\(D=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
Vì \(xyz=2017\)
\(D=\dfrac{xy\left(xz\right)}{xy\left(1+xz+z\right)}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz+1+z}{1+xz+z}=1\)
Vậy D = 1

thay xyz=2017 vaf 2017=xyz a đc :

\(\frac{xyz.x}{xy+xyz.x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)=\(\frac{xyz.x}{xy.\left(xz+z+1\right)}+\frac{y}{y.\left(xz+z+1\right)}+\frac{z}{xz+z+1}\)

=\(\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=\frac{xz+z+1}{xz+z+1}=1\)

Nhanh k cho nè

làm lần lượt nhá,dài dòng quá khó coi.ahihihi!

\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)

Ta có : A = \(\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{xz+z+1}\)

A = \(\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\) (Vì xyz = 2017)

A = \(\dfrac{xy\left(xz\right)}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)

A = \(\dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}\)

A = \(\dfrac{xz+1+z}{xz+1+z}\) = 1

Vậy A = 1

Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)

  \(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)

Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)

                         \(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

Cộng vế với vế ta có:

\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)

\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)

Tại sao x=y=z=$\sqrt{\dfrac{2017}{3}}$ vậy ạ?

Lời giải:

Ta có: \(x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0\)

\(\Leftrightarrow \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=0\)

\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)

Vì \((x-y)^2; (y-z)^2;(z-x)^2\geq 0\), do đó để tổng của chúng bằng $0$ thì:

\((x-y)^2=(y-z)^2=(z-x)^2=0\Rightarrow x=y=z\)

\(\Rightarrow 3x^{2017}=3y^{2017}=3z^{2017}=x^{2017}+y^{2017}+z^{2017}=9\)

\(\Rightarrow x=y=z=\sqrt[2017]{3}\)

\(\Rightarrow \left(\frac{2017x+2018y-4023z}{3}\right)^{2017}=\left(\frac{12x}{3}\right)^{2017}=(4x)^{2017}=3.4^{2017}\)

Em cảm ơn cô chúc cô ngày nhà giáo vui vẻ

\(\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\\ \Leftrightarrow\left(x+y+z\right)\left(xy+yz+zx\right)-xyz=0\\ \Leftrightarrow3\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

\(\forall x=-y\Leftrightarrow VT=-y^{2017}+y^{2017}+z^{2017}=z^{2017}=\left(-y+y+z\right)^{2017}=VP\\ \forall y=-z\Leftrightarrow VT=x^{2017}-z^{2017}+z^{2017}=x^{2017}=\left(x-z+z\right)^{2017}=VP\\ \forall z=-x\Leftrightarrow VT=x^{2017}+y^{2017}-x^{2017}=y^{2017}=\left(x+y-x\right)^{2017}=VP\)

Vậy ta đc đpcm