Tính
B = 2 mũ 20 - 2 mũ 19 - 2 mũ 18 - 2 mũ 17 -...- 2 mũ 1 - 2 mũ 0
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B = 2^20 - ( 2^19 + 2^18 + 2^17 + ... + 2^1 + 2^0
B = (2^19 x 2 - 2^19) - 2^18 - 2^17 - ... - 2^1 - 0
B = (2^18 x 2 - 2^18) - 2^17 - 2^16 - ... - 2^1 - 0
B = (2^17 x 2 - 2^17) - 2^16 - 2^15 - ... - 2^1 - 0
Cứ tách xong lại đóng ngoặc cho đến:
B = 2^1 x 2 - 2^1 - 0
B = 2^1 - 0
B = 2
Đặt A=219+218+...+20
2A=2(219+218+...+1)
2A=220+219+...+2
2A-A=(220+219+...+2)-(219+218+...+2)
A=220-2
Thay A vào B ta có: 220-(220-2)
=220-220+2
=0+2=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
a,\(5^3.2-100:4+2^3.5\)
= 125 . 2 - 25 + 8 . 5
= 250 - 25 + 40
= 265
b, \(6^2:9+50.2-3^3.3\)
= 36 : 9 + 100 - 27 . 3
= 4 + 100 - 81
= 23
a: \(=\left\{145-\left[130-10\right]:2\right\}\cdot5\)
\(=\left\{145-60\right\}\cdot5=85\cdot5=425\)
b: \(=100:\left\{250:\left[450-4\cdot125+4\cdot25\right]\right\}\)
\(=\dfrac{100}{250:\left[450-500+100\right]}=\dfrac{100}{250:50}=\dfrac{100}{5}=20\)
c: \(=355-5\cdot\left[64-\left(27-25\right)\right]=355-5\cdot\left[64-2\right]\)
\(=355-310=45\)
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
Kết quả
= 2^18 (2^5+2^6+2^7)/2^18 (2^0 + 2^1+2^2) = 2^5 (2^0+2^1+2^2)/(2^0 + 2^1+2^2) = 2^5 = 32
Ta có: 2B = \(2^{21}-2^{20}-...-2^0\)
B = \(2^{20}-2^{19}-...-2^1\)
=> 2B + B = \(2^{21}-1\)
=> 3B = \(2^{21}-1\)
=> B = \(\frac{2^{21}-1}{3}\)
2^20 - 2^19 - 2^18 - 2^17 - ... - 2^1 - 2^0
=(2^19 x 2 - 2^19) - 2^18 - 2^17 - 2^16 - ... - 2^1
=(2^18 x 2 - 2^18) - 2^17 - 2^16 - 2^15 - ... - 2^1
Cứ tính xong trong ngoặc lại ghép tiếp........
=2^1 x 2 - 2^1
=2^1
=2