ta thấy \(\begin{cases}\left(2x-5\right)^{2000}\\\left(3y+4\right)^{2002}\end{cases}\ge0}\)  

Theo bài ra ta có (2x-5)²⁰⁰⁰+(3y+4)²⁰⁰²\(\le\) 0

=> (2x-5)²⁰⁰⁰+(3y+4)²⁰⁰²=0

=>2x-5=0 => x=2,5

=>3y+4=0=>y=\(\frac{-4}{3}\)

Vì (2x-5)²⁰⁰⁰ > 0 với mọi x

(3y+4)²⁰⁰² > 0 với mọi y

=>(2x-5)²⁰⁰⁰+(3y+4)²⁰⁰² > 0 ới mọi x;y

Mà (2x-5)²⁰⁰⁰+(3y+4)²⁰⁰² < 0 (theo đề)

=>(2x-5)²⁰⁰⁰+(3y+4)²⁰⁰²=0

=>(2x-5)²⁰⁰⁰=(3y+4)²⁰⁰²=0

+)(2x-5)²⁰⁰⁰=0=>2x-5=0=>x=5/2

+)(3y+4)²⁰⁰²=0=>3y+4=0=>y=-4/3

Vậy x=5/2;y=-4/3

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

\(\left(2x-5\right)^{2000}\ge0\forall x;\left(3y+4\right)^{2002}\ge0\forall y\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2000}\ge0\forall x,y\)

Kết hợp giả thiết ta có:\(2x-5=0;3y+4=0\Rightarrow x=\frac{5}{2};y=-\frac{4}{3}\)

(2x - 5)²⁰⁰⁰ + (3y + 4)²⁰⁰²

ta có: (2x - 5)2000 \(\ge\) 0 ; (3y + 4)2002 \(\ge\) 0

=> (2x - 5)2000 + (3y + 4)2002 \(\ge\) 0

Dấu "=" xảy ra khi 2x - 5 = 0  và 3y + 4 = 0

=> 2x = 5 và 3y = -4

=> x = 2,5 và y = \(\frac{-4}{3}\)

bé hơn mà có phải lớn hơn 0 đâu ?

Do (2x-5)²⁰⁰⁰>0

(3y+4)²⁰⁰²>0

Mà (2x-5)²⁰⁰⁰+(3y+4)²⁰⁰²<0

=>(2x-5)²⁰⁰⁰=0 (3y+4)²⁰⁰²=0

<=>x=2,5 y=4/3

Ta có :

\(\left(2x-5\right)^{2000}\ge0\forall x\)

\(\left(3y+4\right)^{2002}\ge0\forall y\)

\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\forall x;y\)

Mà theo GT : \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)

\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}=0\)

Dấu \("="\) xảy ra

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2000}=0\\\left(3y+4\right)^{2002}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\)

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

Vậy \(x=\dfrac{5}{2};y=-\dfrac{4}{3}\)

(2x-4)²⁰⁰⁰+(3y-10)²<0

Mà (2x-4)²⁰⁰⁰>0 ; (3y-10)²>0

=>(2x-4)²⁰⁰⁰+(3y-10)²<0

<=>(2x-4)²⁰⁰⁰=0;(3y-10)=0

<=>x=2;y=10/3

=>x+y=2+10/3=16/3

Ta có: \(\left(2x-8\right)^{2000}+\left(3y+4\right)^{2022}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-8=0\\3y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=8\\3y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-\dfrac{4}{3}\end{matrix}\right.\)