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

\(\Leftrightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

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

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...

 [2x-5]^2016+[3y+4]^2014<hoặc=0

=>2x-5=0 và 3y+4=0 (vì  [2x-5]^2016+[3y+4]^2014>hoặc=0 với mọi x;y)

=>x=5/2 và y=-4/3

vậy x=5/2 và y=-4/3

thank you

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

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❤

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

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!

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