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\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\2y+5=0\\4z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)
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 \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x-5=0\\2y+5=0\\4z-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)
Vì: \(\left|3x-5\right|\ge0\)và: \(\left(2y+5\right)^{208}\ge0\)cùng với: \(\left(4z-3\right)^{20}\ge0\)
\(\Rightarrow\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}\ge0\)( trái với đề bài )
\(\Rightarrow\)Không tồn tại \(x,y,z\)thỏa mãn đề bài
Chúc bạn học tốt !
Có: \(\left|3x-5\right|\ge0\)
\(\left(2y+5\right)^{208}\ge0\)
\(\left(4z-3\right)^{20}\ge0\)
=> \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}\ge0\)với mọi x, y, z. (1)
Đề bài \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}\le0\) (2)
Từ (1) và (2) Suy ra chỉ xảy ra trường hợp: \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}=0\)
<=> \(3x-5=0;2y+5=0;4z-3=0\)
<=> x =5/3; y=-5/2; z =3/4
\(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}\le0\)
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 \(x=\dfrac{5}{3};y=-\dfrac{5}{2};z=\dfrac{3}{4}\)
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 \(x=\dfrac{5}{3};y=-\dfrac{5}{2};z=\dfrac{3}{4}\)
\(\left(2x-1\right)^2+\left(y-3\right)^8+\left(z-5\right)^{20}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\\z-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\\z=5\end{matrix}\right.\)
\(\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}\le0\)
ta có:
\(\hept{\begin{cases}\left|3x-5\right|\ge0\\\left(2y+8\right)^{20}\ge0\\\left(4z-3\right)^{2018}\ge0\end{cases}}\Rightarrow\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}\ge0\)
mà \(\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}\le0\)=> \(\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}=0\)
=> \(\hept{\begin{cases}\left|3x-5\right|=0\\\left(2y+8\right)^{20}=0\\\left(4z-3\right)^{2018}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-4\\z=\frac{4}{3}\end{cases}}\)
vậy \(x=\frac{5}{3},y=-4,z=\frac{4}{3}\)
bạn nên có một bước giải thích vì sao
(2y+8)\(\ge0\)
\(|3x-5|\ge0\)
\((4z-3)\ge0\)
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 \(x=\frac{5}{3};y=\frac{-2}{5};z=\frac{3}{4}\)