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\(\left(3x-\frac{5}{9}\right)^{2002}+\left(3y+\frac{0,4}{3}\right)^{2004}=0\)
Ta thấy \(\left(3x-\frac{5}{9}\right)^{2002}\ge0\text{ với mọi x}\\ \left(3y+\frac{0,4}{3}\right)^{2004}\ge0\text{ với mọi y}\)
Mà \(\left(3x-\frac{5}{9}\right)^{2002}+\left(3y+\frac{0,4}{3}\right)^{2004}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(3x-\frac{5}{9}\right)^{2002}=0\\\left(3y+\frac{0,4}{3}\right)^{2004}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-\frac{5}{9}=0\\3y+\frac{0,4}{3}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}3x=\frac{5}{9}\\3y=\frac{-0,4}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{\frac{5}{9}}{3}\\y=\frac{\frac{-0,4}{3}}{3}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\frac{5}{27}\\y=\frac{-2}{45}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\frac{5}{27};\frac{-2}{45}\right)\)
\(\left(3x-\frac{5}{9}\right)^{2002}+\left(3y+\frac{0,4}{4}\right)^{2004}=0\)
Ta có: \(\left\{{}\begin{matrix}\left(3x-\frac{5}{9}\right)^{2002}\ge0;\forall x,y\\\left(3y+\frac{0,4}{3}\right)^{2004}\ge0;\forall x,y\end{matrix}\right.\)\(\Rightarrow\left(3x-\frac{5}{9}\right)^{2002}+\left(3y+\frac{0,4}{4}\right)^{2004}\ge0;\forall x,y\)
Do đó \(\left(3x-\frac{5}{9}\right)^{2002}+\left(3y+\frac{0,4}{4}\right)^{2004}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-\frac{5}{9}\right)^{2002}=0\\\left(3y+\frac{0,4}{3}\right)^{2004}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-\frac{5}{9}=0\\3y+\frac{0,4}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{27}\\y=\frac{-2}{45}\end{matrix}\right.\)
Vậy ...
( 3x-5 /9 )^2002 > 0 ; ( 3y+0,4/3 )^2004 > 0
=> (3x-5/9 )^2002 = 0 và ( 3y + 0,4 / 3 )^2004 = 0
=> 3x - 5 = 0
3x = 5
x = 5/3
=> 3y + 0,4 = 0
3y = -0,4
y= -2/15
\(\left(\dfrac{3x-5}{9}\right)^{2018}>=0\forall x\)
\(\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall y\)
Do đó: \(\left(\dfrac{3x-5}{9}\right)^{2018}+\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3x-5}{9}=0\\\dfrac{3y+0,4}{3}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-5=0\\3y+0,4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{0.4}{3}=-\dfrac{2}{15}\end{matrix}\right.\)
