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

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

Nên:  \(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\le0\)

\(\Leftrightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}=0\)

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

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

Ta có: \(\left|3x-5\right|\ge0;\left(2y+5\right)^{2018}\ge0;\left(4z-3\right)^{2020}\ge0.\)mọi x,y, z.

=> \(\left|3x-5\right|+(2y+5)^{2018}+\left(4z-3\right)^{2020}\ge0\)với mọi x, y,z.

Như vậy (1) chỉ xảy ra trường hợp: \(\left|3x-5\right|+(2y+5)^{2018}+\left(4z-3\right)^{2020}=0\)

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

Vậy...

thầy mình cho đè kia cơ

Nếu đề đúng là như vậy thì làm như sau :

                                                    Bài giải

Vì :  \(\left|3x-5+\left(2y+5\right)^{2018}+\left(4z-3\right)^{2020}\right|\ge0\)

\(\Rightarrow\) Chỉ xảy ra trường hợp :

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

Mà \(\hept{\begin{cases}\left(2y+5\right)^{2018}\ge0\\\left(4z-3\right)^{2020}\ge0\end{cases}}\)  \(\Rightarrow\hept{\begin{cases}3x-5=0\\\left(2y+5\right)^{2018}=0\\\left(4z-3\right)^{2020}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}3x-5=0\\2y+5=0\\4z-3=0\end{cases}}\)

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

\(\Rightarrow\text{ }x=\frac{5}{3}\text{ ; }y=-\frac{5}{2}\text{ ; }z=\frac{3}{4}\)

(2x-y+7)^2022>=0 với mọi x,y

|x-3|^2023>=0 với mọi x,y

Do đó: (2x-y+7)^2022+|x-3|^2023>=0 với mọi x,y

mà \(\left(2x-y+7\right)^{2022}+\left|x-3\right|^{2023}< =0\)

nên \(\left(2x-y+7\right)^{2022}+\left|x-3\right|^{2023}=0\)

=>2x-y+7=0 và x-3=0

=>x=3 và y=2x+7=2*3+7=13

a,x=2

b,x=5

x=4

x=6

a) \(\left(2x+1\right)^3=125\)

\(\Rightarrow\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=5-1\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

Vậy x = 2

b) \(\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)

\(\Rightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2=1\end{cases}}\)

TH 1 : \(\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\)

TH 2 : \(\left(x-5\right)^2=1\Rightarrow\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

Vậy  \(x\in\left\{5;6;4\right\}\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{cases}}\)

TH 1 : \(\left(2x-15\right)^3=0\Rightarrow2x-15=0\Rightarrow2x=15\Rightarrow x=\frac{15}{2}\)

TH 2 : \(\left(2x-15\right)^2=1\Rightarrow\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}\)

Vậy  \(x\in\left\{\frac{15}{2};8;7\right\}\)

_Chúc bạn học tốt_

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

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

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

Khi đó thay vào ta được: 

\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)

\(\Rightarrow M=-\frac{1159}{36}\)