Sửa đề: \(\dfrac{x-4}{2019}+\dfrac{x-3}{2018}=\dfrac{x-2}{2017}+\dfrac{x-1}{2016}\)

\(\Leftrightarrow\dfrac{x-4}{2019}+1+\dfrac{x-3}{2018}+1=\dfrac{x-2}{2017}+1+\dfrac{x-1}{2016}+1\)

\(\Leftrightarrow\dfrac{x+2015}{2019}+\dfrac{x+2015}{2018}=\dfrac{x+2015}{2017}+\dfrac{x+2015}{2016}\)

\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x=-2015\) vì \(\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)\ne0\)

đoạn cuối cùng mk chưa hiểu lắm

Lời giải:

Từ điều kiện đề bài suy ra:

\(\left\{\begin{matrix} x^{2016}+y^{2016}-x^{2017}-y^{2017}=0\\ x^{2017}+y^{2017}-x^{2018}-y^{2018}=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x^{2016}(1-x)+y^{2016}(1-y)=0\\ x^{2017}(1-x)+y^{2017}(1-y)=0\end{matrix}\right.\)

\(\Rightarrow x^{2016}(1-x)(1-x)+y^{2016}(1-y)(1-y)=0\) (trử theo vế)

\(\Leftrightarrow x^{2016}(1-x)^2+y^{2016}(1-y)^2=0\)

Dễ thấy \(x^{2016}(1-x)^2; y^{2016}(1-y)^2\geq 0\) nên để tổng của chúng bằng $0$ thì:
\(x^{2016}(1-x)^2=y^{2016}(1-y)^2=0\)

\(\Rightarrow (x,y)=(0,1), (0,0), (1,1)\) và hoán vị của nó

Thử lại vào đk ban đầu thấy thỏa mãn

Do đó: \(A=x^{2019}+y^{2019}\in\left\{0; 1;2\right\}\)

Vì \(x^{2016}+y^{2016}=x^{2017}+y^{2017}=x^{2018}+y^{2018}\left(x,y\ge0\right)\)

\(\Rightarrow x=y=1\)

\(\Rightarrow A=1^{2019}+1^{2019}\)

\(\Rightarrow A=2\)

\(\frac{x-1}{2017}+\frac{x-2}{2018}+\frac{x-3}{2019}=-3\)

\(\Leftrightarrow\frac{x-1}{2017}+1+\frac{x-2}{2018}+1+\frac{x-3}{2019}+1=0\)

\(\Leftrightarrow\frac{x+2016}{2017}+\frac{x+2016}{2018}+\frac{x+2016}{2019}=0\)

\(\left(x+2016\right)\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)=0\)

vì \(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\ne0\) nên

x+2016=0

\(\Leftrightarrow\)x=-2016

\(\dfrac{2-x}{2017}+1=\dfrac{x-1}{2018}-1+1-\dfrac{x}{2019}\)

\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{x-2019}{2018}+\dfrac{2019-x}{2019}\)

\(\Leftrightarrow\dfrac{2019-x}{2017}+\dfrac{2019-x}{2018}-\dfrac{2019-x}{2019}=0\)

\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)

\(\Leftrightarrow2019-x=0\) (do \(\dfrac{1}{2017}>\dfrac{1}{2019}\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}>0\))

\(\Rightarrow x=2019\)

Ta có \(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}.x=\frac{2018}{2019}.x\)

<=>\(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}x-\frac{2018}{2019}x=0\)

<=>x\(\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)

Vì \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\) không thể bằng 0

Vậy x=0

Ta có 1 nghiệm thỏa mãn S=\(\left\{0\right\}\)

\(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\Leftrightarrow\left(\dfrac{2-x}{2017}+1\right)=\left(\dfrac{1-x}{2018}+1\right)+\left(1-\dfrac{x}{2019}\right)\)

\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{2019-x}{2018}+\dfrac{2019-x}{2019}\)\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)

Ta đã có: \(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}< 0\)

Vậy ta dễ dàng suy ra được \(S=\left\{2019\right\}\)

Cố gắng học Thư nhé! Say oh yeah!

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\\z-\frac{1}{z}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\y^2=1\\z^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm1\\z=\pm1\end{matrix}\right.\)

Vậy P có thể nhận các giá trị \(P=\left\{-1;1;3\right\}\)