CỨU TUI VỚI GẤP
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:
\(\dfrac{31}{23}-\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\)
\(=\dfrac{31}{23}-\dfrac{7}{32}-\dfrac{8}{23}\)
\(=\left(\dfrac{31}{23}-\dfrac{8}{23}\right)-\dfrac{7}{32}\)
\(=\dfrac{23}{23}-\dfrac{7}{32}\)
\(=1-\dfrac{7}{32}\)
\(=\dfrac{32}{32}-\dfrac{7}{32}\)
\(=\dfrac{25}{32}\)
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`