Biến đổi:\(\left(\dfrac{3x-5}{9}\right)^{2014}+\left(\dfrac{3y+0,4}{3}\right)^{2016}=\dfrac{\left(3x-5\right)^{2014}}{9^{2014}}+\dfrac{\left(3y+0,4\right)^{2016}}{9^{1008}}=\dfrac{\left(3x-5\right)^{2014}+9^{1006}\left(3y+0,4\right)^{2016}}{9^{2016}}\)
=>\(\left(3x-5\right)^{2014}+9^{1006}\left(3y+0,4\right)^{2016}=0\)
Do x;y nguyên
=>(3x-5)2014 là 1 số nguyên
91006(3y+0,4)2016 là số thập phân
=>tổng của chúng khác 0
=>không tồn tại x;y thõa mãn
mk đây :v
Ta có :
\(\left(\dfrac{3x-5}{9}\right)^{2014}+\left(\dfrac{3y+0,4}{3}\right)^{2016}=0\)
Mà :
\(\left\{{}\begin{matrix}\left(\dfrac{3x-5}{9}\right)^{2014}\ge0\\\left(\dfrac{3y+0,4}{3}\right)^{2016}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{3x-5}{9}\right)^{2014}=0\\\left(\dfrac{3y+0,4}{3}\right)^{2016}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x-5}{9}=0\\\dfrac{3y+0,4}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\3y+0,4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=5\\3y=-0,4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{-0,4}{3}\end{matrix}\right.\)
Vậy .......................
\(\left(\dfrac{3x-5}{9}\right)^{2014}+\left(\dfrac{3y+0,4}{3}\right)^{2016}=0\)
\(\left\{{}\begin{matrix}\left(\dfrac{3x-5}{9}\right)^{2014}\ge0\\\left(\dfrac{3y+0,4}{3}\right)\ge0\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{3x-5}{9}\right)^{2014}+\left(\dfrac{3y+0,4}{3}\right)^{2016}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(\dfrac{3x-5}{9}\right)^{2014}=0\\\left(\dfrac{3y+0,4}{3}\right)^{2016}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-5}{9}=0\Rightarrow3x-5=0\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\\\dfrac{3y+0,4}{3}=0\Rightarrow3y+0,4=0\Rightarrow3y=-0,4\Rightarrow y=\dfrac{-0,4}{3}\end{matrix}\right.\)
\(\)
