\(\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}=0\)

Ta có : \(\hept{\begin{cases}\left(\frac{3x-5}{9}\right)^{2018}\ge0\forall x\\\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall y\end{cases}}\Rightarrow\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{3x-5}{9}=0\\\frac{3y+0,4}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x-5=0\\3y+0,4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{2}{15}\end{cases}}\)

( 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

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 .......................

Các bạn đâu rồi?????

\(\left(\dfrac{3x-5}{9}\right)^{2018}>=0\forall x\)

\(\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall y\)

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 \(\left\{{}\begin{matrix}\dfrac{3x-5}{9}=0\\\dfrac{3y+0,4}{3}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-5=0\\3y+0,4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{0.4}{3}=-\dfrac{2}{15}\end{matrix}\right.\)

\(\left(3x-\frac{5}{9}\right)^{2002}+\left(3y+\frac{0,4}{3}\right)^{2004}=0\)

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 ...

\(2x:15=0,4:0,8\)

\(=2x:15=0,5\)

\(\Rightarrow2x=7,5:2\)

\(\Rightarrow x=3,75\)

Mk pt nhiu đóa à!

4.(3x-2)-10=4(x-2)
=>4.(3x-2)-10-4.(x-2)=0
=> 4.[ (3x-2)-(x-2)] -10=0
=> 4.2x-10=0
=>8x-10=0
=>8x=10
=>x= 5/4
 

=>2x:15=0,5

=>2x=7,5

=>x=3,75