Tham Khảo:

https://olm.vn/hoi-dap/detail/62824771750.html

+) 9,65 x 0,4 x 2,5

= 9,65 x ( 0,4 x 2,5 )

= 9,65 x 1

= 9,65

+) 7,38 x 1,25 x 80

= 7,38 x (1,25 x 80)

= 7,38 x 100

= 738

+) 0,25 x 40 x 9,84

= 10 x 9,84

= 98,4

+) 34,3 x 5 x 0,4

= 34,3 x 2

= 68,6

9,65 x 0,4 x 2,5

= 9,65 x 1

= 9,65

7,38 x 1,25 x 80

= 7,38 x 100

= 738

0,25 x 40 x 9,84

= 9,84 x 10

= 98,4

34,3 x 5 x 0,4

= 34,3 x 2

= 68,6

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

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

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

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

`@Tham` `Khảo`

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

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 ...

Có:

\(0,4-\dfrac{3}{4}.\dfrac{-5}{9}\)

\(=\dfrac{2}{5}-\dfrac{3}{4}.\dfrac{-5}{9}\)

\(=\dfrac{2}{5}-\dfrac{-5}{12}\)

\(=\dfrac{2}{5}+\dfrac{5}{12}=\dfrac{49}{60}\)

Chúc bạn học tốt!

\(0,4-\dfrac{3}{4}.\dfrac{-5}{9}=\dfrac{2}{5}-\dfrac{-5}{12}=\dfrac{2}{5}+\dfrac{5}{12}=\dfrac{24}{60}+\dfrac{25}{60}=\dfrac{49}{60}\)

Mấy cái này dễ lắm nên lần sau bạn tự làm đi nha

\(0,4-\dfrac{3}{4}.\dfrac{-5}{9}=\dfrac{2}{5}+\dfrac{5}{12}=\dfrac{49}{60}\)

Cái này bạn dùng máy tính là ra mà.

Chúc bạn học tốt!!!

b) \(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}\)

\(=\dfrac{\left(0,8\right)^5}{\left(0.4\right)^5\cdot0,4}\)

\(=\left(\dfrac{0,8}{0,4}\right)^5\cdot\dfrac{1}{0,4}\)

\(=2^5\cdot\dfrac{5}{2}\)

\(=\dfrac{2^5\cdot5}{2}\)

\(=2^4\cdot5\)

\(=80\)

c) \(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}\)

\(=\dfrac{2^{15}\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}\)

\(=\dfrac{2^{15}\cdot3^8}{2^{15}\cdot3^6}\)

\(=\dfrac{3^2}{1}\)

\(=9\)

c) 2^15.9^466.83=2^15.3^8/2^6.3^6.2^9=3^2/1=9

dấu này / là phần  nha 

1&2/3 là gì zậy ta =)?

a) 5/3

b)51/35