\(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}\le0\)
Ta có:
\(\left|3x-5\right|\ge0\)
\(\left(2y+5\right)^{208}\ge0\)
\(\left(4z-3\right)^{20}\ge0\)
\(\Rightarrow\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}\ge0\)
\(\Rightarrow\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x-5\right|=0\\\left(2y+5\right)^{208}=0 \\\left(4z-3\right)^{20}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x-5=0\\2y+5=0\\4z-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=5\\2y=-5\\4z=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(x=\dfrac{5}{3};y=-\dfrac{5}{2};z=\dfrac{3}{4}\)

la

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

Sửa đề \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4x-3\right)^{20}\le0\)

Mà \(\left|3x-5\right|\ge0\);\(\left(2y+5\right)^{208}\ge0;\left(4x-3\right)^{20}\ge0\)

Do đó \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}=0\)

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

Ta có: \(\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}\ge0\)với \(\forall x;y;z\)

Mà \(\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}\le0\)

\(\Rightarrow\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}=0\)

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

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

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

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

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

Ta có: \(\left|3x-5\right|\ge0\forall x\)

\(\left(2y+5\right)^{20}\ge0\forall y\)

\(\left(4z-3\right)^{206}\ge0\forall z\)

Do đó: \(\left|3x-5\right|+\left(2y+5\right)^{20}+\left(4z-3\right)^{206}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{3};y=-\dfrac{5}{2};z=\dfrac{3}{4}\)

x, y, z thuộc j

câu trả lời là : 80

×=5/2 y=-5/2 z = 3/4