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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:
\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\)
\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)
______
\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)
\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)
_____
\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)
\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)
a)64:2mũ5×30×4
= 64 : 32 x 30 x 4
= 240
b)3 mũ 2× 5 - 2 mũ 2×7+2 mũ 0 × 5
= 9 x 5 - 4 x 7 + 1 x 5
= 45 - 28 + 5
= 22
c)2 mũ 3-5 mũ 3÷5 mũ 2 + 12×2 mũ 2
= 8 - 125 : 25 + 12 x 4
= 8 - 5 + 48
= 51
d)2[(7-3 mũ 3÷3 mũ 2) chia 2 mũ 2 + 99]-100
= 2[( 7 - 27 : 9) : 4 + 99] - 100
= 2[4 : 4 + 99] - 100
= 2. 100 - 100
= 200 - 100
= 100
e)4[(3 + 3^7:3^4)chia 10 + 97]-300
= 4[( 3 + 3^3) : 10 + 97] - 300
= 4[ 30 : 10 + 97 ] - 300
= 4. 100 - 300
= 400 - 300
= 100
f)2^2 x 5 [(5 mũ 2 cộng 2 mũ 3) chia 11 - 2] - 3^2 x 2
= 4 x 5 [ (25 + 8 ) : 11 - 2] - 9 x 2
= 20 [ 33 : 11 - 2] - 18
= 20. 1 - 18
= 20 - 18
= 2
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a: \(3^x-2=2^7\)
\(\Leftrightarrow3^x=128+2=130\)(vô lý)
b: \(4^{x+1}=64\)
=>x+1=3
hay x=2
c: \(\left(5x+1\right)^2=1^{2016}=1\)
=>5x+1=1 hoặc 5x+1=-1
=>x=0 hoặc x=-2/5
d: \(2^{2\left(x-1\right)}=8\)
=>2(x-1)=3
=>x-1=3/2
hay x=5/2
1. 53 = 5.5.5 = 125
2. 27 = 2.2.2.2.2.2.2 = 128
3. 44 = 4.4.4.4 = 256
4. 73 = 7.7.7 = 343
6. 35 = 243
7. 26 = 64
8. 34 = 81
9. 83 = 512
11. 132 = 169
12. 112 = 121
13. 142 = 196
14. 152 = 225
16. 172 = 289
17. 182 = 324
18. 192 = 361
19. 202 = 400
21. 104 = 10000
22. 105 = 100000
23. 106 = 1000000
24. 107 = 10000000
b=22x27:24
b=22+7-4
b=25=32
\(B=2^2\cdot2^7:2^4\)
\(B=2^{2+7-4}\)
\(B=2^{2+3}\)
\(B=2^5\)
\(B=32\)