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\(\left(2x-1\right)^2+\left(y-3\right)^8+\left(z-5\right)^{20}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\\z-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\\z=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\2y+5=0\\4z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)
\(\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}\le0\)
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: \(\left|3x-5\right|\ge0\forall x\)
\(\left(2y+5\right)^{20}\ge0\forall y\)
\(\left(4z-3\right)^{206}\ge0\forall z\)
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 \(x=\dfrac{5}{3};y=-\dfrac{5}{2};z=\dfrac{3}{4}\)
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 \(x=\frac{5}{3};y=\frac{-2}{5};z=\frac{3}{4}\)
Xét \(\left|3x-5\right|\ge0\)
\(\left(2y+5\right)^{20}\ge0\)
\(\left(4z-3\right)^{206}\ge0\)
\(\Rightarrow\left|3x-5\right|+\left(2y+5\right)^{20}+\left(4z-3\right)^{206}\ge0\)(1)
Mà: \(\left|3x-5\right|+\left(2y+5\right)^{20}+\left(4z-3\right)^{206}\le0\)(2)
(1)(2) suy ra: \(\left|3x-5\right|+\left(2y+5\right)^{20}+\left(4z-3\right)^{206}=0\)
\(\hept{\begin{cases}3x-5=0\Rightarrow3x=5\Rightarrow x=\frac{5}{3}\\\left(2y+5\right)^{20}=0\Rightarrow2y+5=0\Rightarrow2y=-5\Rightarrow y=-\frac{5}{2}\\\left(4z-3\right)^{206}=0\Rightarrow4z-3=0\Rightarrow4z=3\Rightarrow z=\frac{3}{4}\end{cases}}\)
Vậy............