Bạn vào câu hỏi tương tự có nha, hoặc vào link này: https://hoc24.vn/hoi-dap/question/198034.html

\(\dfrac{x}{2015}=\dfrac{y}{2016}=\dfrac{z}{2017}=\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-1}\Leftrightarrow x-z=x-y=y-z\Rightarrow x=y=z.\)

vãi cái đề

Áp dụng tc dtsbn:

\(\dfrac{x}{2013}=\dfrac{y}{2014}=\dfrac{z}{2015}=\dfrac{x-z}{-2}=\dfrac{y-z}{-1}=\dfrac{x-y}{-1}\\ \Leftrightarrow\dfrac{x-z}{2}=\dfrac{y-z}{1}=\dfrac{x-y}{1}\\ \Leftrightarrow x-z=2\left(y-z\right)=2\left(x-y\right)\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

\(x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)

Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)

Tương tự:

\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)

\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)

\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)