Question

CỨU TUI VỚI GẤP 

Answer 1

Bài 2:
\(a,\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}\)
\(=\left(\dfrac{2}{3}\right)^{2+1}\)
\(=\left(\dfrac{2}{3}\right)^3\)
\(b,\left(\dfrac{-4}{9}\right)^5:\dfrac{16}{81}\)
\(=\left(\dfrac{-4}{9}\right)^5:\dfrac{4^2}{9^2}\)
\(=\left(\dfrac{-4}{9}\right)^5:\left(\dfrac{-4}{9}\right)^2\)
\(=\left(\dfrac{-4}{9}\right)^{5-2}\)
\(=\left(\dfrac{-4}{9}\right)^3\)
\(c,\left[\left(\dfrac{1}{2}\right)^3\right]^2\)
\(=\left(\dfrac{1}{2}\right)^6\)
Bài 3:
\(\dfrac{31}{23}-\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\)
\(=\dfrac{31}{23}-\dfrac{7}{32}-\dfrac{8}{23}\)
\(=\left(\dfrac{31}{23}-\dfrac{8}{23}\right)-\dfrac{7}{32}\)
\(=\dfrac{23}{23}-\dfrac{7}{32}\)
\(=1-\dfrac{7}{32}\)
\(=\dfrac{32}{32}-\dfrac{7}{32}\)
\(=\dfrac{25}{32}\)

Answer 2

Bài 1:

có \(\sqrt{2}\approx1,42\)

`1,(4)= 1,44...

`=>`\(-1< 1,41< 1,44..< 2\)

Sắp xếp theo thứ tự tăng dần là :

\(-1;\sqrt{2};1,\left(4\right)2\)

Bài 3:

`31/23 - (7/32 + 8/23)`

`= 31/23 - 7/32 - 8/23`

`=31/23 - 8/23 - 7/32`

`= 23/23 - 7/32`

`= 1 - 7/32`

`= 32/32 - 7/32`

`= 25/32`