2^4
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`18/27-8/27`
`=10/27`
`4/7xx2/5`
`=8/35`
`3/4:2/7`
`=3/4xx7/2`
`=21/8`

Ta có công thức tổng quát như sau:
\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)
Áp dụng ta có:
\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\)
\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)
______
\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)
\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)
_____
\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)
\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)

(1 - 2)2 + (3 - 4)3 + (4 - 5)4 + ... + (99 - 100)99
= (-1)2 + (-1)3 + (-1)4 + ... + (-1)99
= 1 + (-1) + 1 + ... + (-1)
= (1 - 1) + (1 - 1) + ... + (1 - 1)
= 0

Bài 1:
a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)
= \(\dfrac{29}{10}\)
b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)
= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)
= \(\dfrac{19}{20}\)
c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)
= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)
= \(\dfrac{29}{6}\)
Bài 2:
3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\)
= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)
= \(\dfrac{28}{5}\)
b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)
= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)
= \(\dfrac{43}{34}\)

C1:
a) 2^10: 2^8=1024: 256= 4
b) 4^6 : 4^3 = 4096: 64 = 64
c) 8^5 : 8^4 = 32768 : 4096 = 8
d) 7^4 : 7^4 = 2401: 2401= 1
C2
a) 2^10 : 2^8 = 2^2 = 4
b) 4^6 : 4^3 = 4^3 = 64
c) 8^5 : 8^4 = 8
d) 7^4 : 7^4 = 1
2⁴ = 2.2.2.2 = 16
2\(^4\) = 2.2.2.2 = 4.2.2 = 8.2 = 16