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

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

bai 2: a) \(2^{30}=\left(2^3\right)^{10}=8^{10}\)

            \(3^{20}=\left(3^2\right)^{10}=9^{10}\)

vi 8¹⁰ <9¹⁰ nen 2³⁰ <3²⁰

       b)       \(5^{202}=\left(5^2\right)^{101}=25^{101}\)

                 \(2^{505}=\left(2^5\right)^{101}=32^{101}\)

vi 25¹⁰¹ <32¹⁰¹ nen 5²⁰² <2⁵⁰⁵

c) \(333^{444}=\left(3.111\right)^{444}=3^{444}.111^{444}=\left(3^4\right)^{111}.111^{444}=81^{111}.111^{444}\)

   \(444^{333}=\left(4.111\right)^{333}=4^{333}.111^{333}=\left(4^3\right)^{111}.111^{333}=64^{111}.111^{333}\)

vi 81¹¹¹>64¹¹¹ va 111⁴⁴⁴>111³³³

nen 333⁴⁴⁴>444³³³

bai 3 : \(\left(\frac{1}{3}\right)^{2n-1}=3^5\)

 \(\left(\frac{1}{3}\right)^{2n-1}=\left(\frac{1}{3}\right)^{-5}\)

2n-1=-5

2n=-5+1

2n=-4

n=-4:2

n=-2

Bai 4 : 3x-5/9=0 va 3y+0,4/3=0

           3x=5/9 va 3y=2/15

             x=5/27 va y=2/45

Bai 5:

A=75. {4²⁰⁰².(4²+1)+....+(4²+1)+1)+25

A=75.{4²⁰⁰².20+...+20+1}+25

A=75.{20.(4²⁰⁰²+...+1)+1}+25

A=75.20.(4²⁰⁰²+..+1)+75+25

A=1500.(4²⁰⁰²+...+1)+100

A=100.{15.(4²⁰⁰²+...+1)+1} chia het cho 100

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

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

Ta có: \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+\dfrac{2}{5}\right)^{100}\ge0\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+\dfrac{2}{5}\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left(x,y,z\right)=\left(\dfrac{1}{5};\dfrac{-2}{5};3\right)\)

Vì \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0,\left(y+0,4\right)^{100}\ge0,\left(z-3\right)^{678}\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)

Vậy \(\left(x,y,z\right)=\left(\dfrac{1}{5};-0,4;3\right)\)

Vì \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+0,4\right)^{100}\ge\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

\(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\)

mà \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)

Dấu ''='' xảy ra khi \(x=\dfrac{1}{5};y=-0,4;z=3\)