M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2
(2x-5)^2020+(3y+4)^2022<=0

=>x=5/2 và y=-4/3

M=25/4+11*5/2*(-4/3)-16/9=-1159/36

Sửa: \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\le0\)

Mà \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\ge0\) với mọi x,y

Do đó \(\left\{{}\begin{matrix}\dfrac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{1}{18}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=6+18=24\)

Đề thiếu rồi bạn

\(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0\left(1\right)\)

Ta có: \(\hept{\begin{cases}\left(2x-5\right)^{2020}\ge0;\forall x,y\\\left(3y+4\right)^{2018}\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\ge0;\forall x,y\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x-5\right)^{2020}=0\\\left(3y+4\right)^{2018}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{-4}{3}\end{cases}}\)

Vậy...

Vì \(\left(2x-5\right)^{2020}\ge0\forall x\); \(\left(5y+1\right)^{2022}\ge0\forall y\)

\(\Rightarrow\left(2x-5\right)^{2020}+\left(5y+1\right)^{2022}\ge0\forall x,y\)

mà \(\left(2x-5\right)^{2020}+\left(5y+1\right)^{2022}\le0\)( giả thuyết )

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-5=0\\5y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=5\\5y=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{-1}{5}\end{cases}}\)

Vậy \(x=\frac{5}{2}\)và \(y=\frac{-1}{5}\)

( 2x - 5 )²⁰²⁰ + ( 5y + 1 )²⁰²² ≤ 0

Ta có : ( 2x - 5 )²⁰²⁰ ≥ 0 ∀ x

            ( 5y + 1 )²⁰²² ≥ 0 ∀ y

=> ( 2x - 5 )² + ( 5y + 1 )²⁰²² ≥ 0 ∀ x, y

Kết hợp với đề bài => Chỉ xảy ra trường hợp ( 2x - 5 )²⁰²⁰ + ( 5y + 1 )²⁰²² = 0

Khi đó \(\hept{\begin{cases}2x-5=0\\5y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{1}{5}\end{cases}}\)

\(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0\)

Ta có:

\(\left\{{}\begin{matrix}\left(2x-5\right)^{2020}\ge0\\\left(3y+4\right)^{2018}\ge0\end{matrix}\right.\forall xy.\)

\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\ge0\) \(\forall xy.\)

Mà \(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0.\)

\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}=0\)

\(\Rightarrow\left(2x-5\right)+\left(3y+4\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\)

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

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

Chúc bạn học tốt!

giúp mình với

Ta có:

(2x+5)²⁰²⁰ ≥ 0 với ∀ x

(3y+4)²⁰¹⁸ ≥ 0 với ∀ y

⇒ (2x+5)²⁰²⁰ + (3y+4)²⁰¹⁸ ≥ 0 với ∀ x, y

Mà (2x+5)²⁰²⁰ + (3y+4)²⁰¹⁸ ≤ 0

⇒ (2x+5)²⁰²⁰ + (3y+4)²⁰¹⁸ = 0

⇒ \(\left[{}\begin{matrix}\left(2x+5\right)^{2020}=0\\\left(3y+4\right)^{2018}=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x+5=0\\3y+4=0\end{matrix}\right.\)

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

Vậy...

Học tốt❤

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

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

Vì : \(\left(2x-5\right)^{2022}\ge0\forall x,\left(3y+4\right)^{2024}\ge0\forall y\\ =>\left(2x-5\right)^{2022}+\left(3y+4\right)^{2024}\ge0\)

Do đó đề bài xảy ra khi và chỉ khi :

\(\left\{{}\begin{matrix}\left(2x-5\right)^{2022}=0\\\left(3y+4\right)^{2024}=0\end{matrix}\right.\\ =>\left(x;y\right)=\left(\dfrac{5}{2};-\dfrac{4}{3}\right)\)

Mình ko biết cách để làm ra đc kết quả này, có thể giải thích cụ thể hơn ko ạ?

\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)