\(\left(2x+5\right)^{2016}\ge0;\left(5y-4\right)^{2016}\ge0\)

=>\(\left(2x+5\right)^{2016}+\left(5y-4\right)^{2016}\ge0\)

theo đề:\(\left(2x+5\right)^{2016}+\left(5y-4\right)^{2016}\le0\)

=>(2x+5)²⁰¹⁶=(5y-4)²⁰¹⁶=0

ta có:(2x+5)²⁰¹⁶=0=>2x=-5=>x=-5/2=-2,5

(5y-4)²⁰¹⁶=0=>5y=4=>y=4/5=0,8

vậy y=0,8

Ta có: \(\hept{\begin{cases}\left(2x-y-4\right)^{2016}\ge0\\\left(3x+2y-13\right)^{2016}\ge0\end{cases}}\)

\(\Rightarrow\left(2x-y-4\right)^{2016}+\left(3x+2y-13\right)^{2016}\ge0\)

Dấu bằng xảy ra khi

\(\hept{\begin{cases}2x-y-4=0\\3x+2y-13=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}\)

\(\Rightarrow A=\left(x-y\right)^{2016}+2016=\left(3-2\right)^{2016}+2016=2017\)

(3x - 1)^2016 + (5y - 3)^2016 < 0    (1)

có (3x - 1)^2016 > 0 

     (5y - 3)^2018 > 0

=> (3x-1)^2016  + (5y - 3)^2018 > 0    và (1)

=> (3x - 1)^2016 + (5y - 3)^2016 = 0

=> 3x - 1 = 0 và 5y - 3 = 0

=> x = 1/23 và y = 3/5

Thông cảm máy chụp đểu

Ta có: \(\hept{\begin{cases}\left(5x-y\right)^{2016}\ge0\\\left|x^2-4\right|^{2017}\ge0\end{cases}\Rightarrow\left(5x-y\right)^{2016}+\left|x^2-4\right|\ge}0\)

Mà \(\left(5x-y\right)^{2016}+\left|x^2-4\right|^{2017}\le0\)

\(\Rightarrow\hept{\begin{cases}\left(5x-y\right)^{2016}=0\\\left|x^2-4\right|^{2017}=0\end{cases}\Rightarrow\hept{\begin{cases}5x-y=0\\x^2-4=0\end{cases}}\Rightarrow\hept{\begin{cases}y=\pm10\\x=\pm2\end{cases}}}\)

Vậy các cặp (x;y) là (2;10);(-2;-10)

cảm ơn

\(\left|x-1\right|+\left(y+2\right)^{2016}=0\)

Ta thấy: \(\hept{\begin{cases}\left|x-1\right|\ge0\\\left(y+2\right)^{2016}\ge0\end{cases}}\)

\(\Rightarrow\left|x-1\right|+\left(y+2\right)^{2016}\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x-1\right|=0\\\left(y+2\right)^{2016}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

\(\Rightarrow A=2x^5-5y^3+2017=2\cdot1^5-5\cdot\left(-2\right)^3+2017=2059\)

x=5/2,y=-4/3

Vì \(\left(2x-5\right)^{2016}\ge0\forall x;\left(3y+4\right)^{2020}\ge0\forall y\)

\(\Rightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\ge0\)

Mà đề lại cho \(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\le0\)

Nên \(\hept{\begin{cases}\left(2x-5\right)^{2016}=0\\\left(3y+4\right)^{2020}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}}\)

Vậy ..........

vì (2x-5)²⁰¹⁶  và (3y+4)²⁰²⁰ >hoặc=0 với mọi x

=>2x-5=3y+4=0

=>x=2/5;y=-4/3

\(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}\le0\)

Ta có:

\(\left\{{}\begin{matrix}\left(2x-5\right)^{2016}\ge0\\\left(3y+4\right)^{2018}\ge0\end{matrix}\right.\forall x.\)

\(\Rightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-5\right)^{2016}=0\\\left(3y+4\right)^{2016}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=0+5=5\\3y=0-4=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)

Chúc bạn học tốt!