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A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90
2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100
2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - ( 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 )
A = 2^100 - 2^3
B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50
5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51
5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )
4B = 5^51 - 1
B = 5^51 - 1 / 4
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
Hai bài trên áp dụng công thức với khoảng cách là 2.
Ta có:
\(D=1+2^1+2^2+2^3+.....+2^{150}\)
\(\Rightarrow2D-D=\left(2+2^2+2^3+2^4+.....+2^{151}\right)-\left(1+2+2^2+2^3+....+2^{150}\right)\)
\(\Rightarrow D=2^{151}-1\)
\(E=1+4^1+4^2+....+4^{400}\)
\(\Rightarrow4E-E=\left(4+4^2+4^3+....+4^{401}\right)-\left(1+4^1+4^2+....+4^{400}\right)\)
\(\Rightarrow E\left(4-1\right)=4^{401}-1\Leftrightarrow E=\frac{4^{401}-1}{4-1}\)
Các câu còn lại làm tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(2^4\cdot3^5:6^4\)
\(=\frac{2^4\cdot3^6}{\left(2\cdot3\right)^4}\)
\(=\frac{2^4\cdot3^6}{2^4\cdot3^4}\)
\(=3^2\)
Bài 2
\(a,5^3\cdot8=5^3\cdot2^3=10^3=1000\)
\(b,2^5-2019^0=32-1=31\)
\(c,3^3+2^5-1^{10}=27+32-1=58\).
\(d,9^2\cdot33-81\cdot23+5^2=81\cdot33-81\cdot23+25\)
\(=81\cdot\left(33-23\right)+25\)
\(=810+25=835\)
\(g,\left[2^2+6^2\right]:5+11^2\)
\(=\left[4+36\right]:5+121\)
\(=40:5+121=8+121\)
\(=129\)
\(d,\frac{14\cdot3^{10}-5\cdot3^{10}}{3^{12}}\)
\(=\frac{3^{10}\cdot\left(14-5\right)}{3^{12}}\)
\(=\frac{3^{10}\cdot9}{3^{12}}\)
\(=\frac{3^{10}\cdot3^2}{3^{12}}=\frac{3^{12}}{3^{12}}\)
\(=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a=2mu 101 - 2
b= 3 mu 2010 - 1
c=5mu 1999-1
d=4 mu n . 4 -4
a=2+22+...+2100
2a=22+23+24+...+2101
a=2a-a=a
=> a= 22+23+24+..+2101 -(2+2^2+...+2^100)
=>a= 2^101 -2
![](https://rs.olm.vn/images/avt/0.png?1311)
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
B=[(45.79+45.21)]:90-5^2]:5+2^3 B=[(45.79+45.21):90-25]:5+8 B=[(45.(79+21):65]:13 B=[(45.100):65]:13 B=[4500:65]:13 B=4500:65:13
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Giải:
a) Đặt:
\(A=1+2^2+2^3+2^4+...+2^{2018}\)
\(\Leftrightarrow2A=2+2^3+2^4+2^5+...+2^{2019}\)
\(\Leftrightarrow2A-A=\left(2+2^{2019}\right)-\left(1+2^2\right)\)
\(\Leftrightarrow A=2+2^{2019}-1-2^2\)
\(\Leftrightarrow A=2+2^{2019}-5\)
\(\Leftrightarrow A=2^{2019}-3\)
Vậy \(A=2^{2019}-3\).
b) Đặt:
\(B=1+5+5^2+5^3+...+5^{2017}\)
\(\Leftrightarrow5B=5+5^2+5^3+5^4+...+5^{2018}\)
\(\Leftrightarrow5B-B=5^{2018}-1\)
\(\Leftrightarrow4B=5^{2018}-1\)
\(\Leftrightarrow B=\dfrac{5^{2018}-1}{4}\)
Vậy \(B=\dfrac{5^{2018}-1}{4}\).
Chúc bạn học tốt!
a)A= 1 + 22+23 + 24 +....+22018
2A = 22 + 23 + 24 +......+22018 + 22019
_
A= 1 + 22+23 + 24 +....+22018
A= 22019 - 1
\(2^x\div2^5=1\)
\(2^x=2^5\)
x = 5
Vậy x = 5
\(2^x:2^5=1\)
\(=>2^x=1.2^5\)
\(=>2^x=2^5\)
Vậy \(x=5\)