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a) 4 + ( 5x + 2 ) : 3 = 58
( 5x + 2 ) : 3 = 58 - 4
( 5x + 2 ) : 3 = 54
( 5x + 2 ) = 54 . 3
( 5x + 2 ) = 162
5x = 162 - 2
5x = 160
x = 160 : 5
x = 32
a) x = 32.
b) x = 5.
c) a = -1;0;1. Riêng câu này thì mình chứ chắc đứng nha bạn.
giả sử số cần tìm là A , ta có
A=1+2+21+22+....+22019
2A=2+ 21+22+23+....+22020
2A-A= (2+21+22+23+....+22020) - (1+2+21+22+....+22019)
A=22020 - 2
Đăt A= 1+2+22+......+22019
2A=2(1+2+22+.....+22019)
2A=2+22+23+....+22020
2A-A=(2+22+23+.....+22020)-(1+2+22+....+22019)
A =2+22+23+....+22020-1-2-22-....-22019
A=22020-1
21 + 22 + 23 + ... + 2100
Ta có : S = 2 + 22 + 23 + ... + 2100
2S = 2.(2 + 22 + 23 + ... + 2100)
2S = 22 + 23 + ... + 2100 + 2101
2S - S = (22 + 23 + ... + 2100 + 2101) - (2 + 22 + 23 + ... + 2100)
S = 2101 - 2
\(2^1+2^2+2^3+...+2^{100}\)
Ta có : \(S=2+2^2+2^3+....2^{100}\)
: \(2S=2.\left(2+2^2+2^3+....+2^{100}\right)\)
: \(2S=2^2+2^3+.....+2^{100}+2^{101}\)
: \(2S-S=\left(2^2+2^3+....+2^{100}+2^{101}\right)\)\(-\left(2+2^2+2^3+.....+2^{100}\right)\)
: \(S=2^{101}-2\)
a) \(D=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7D=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)
\(\Rightarrow7D-D=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6D=1-\frac{1}{7^{100}}\)
\(\Rightarrow D=\left(1-\frac{1}{7^{100}}\right).\frac{1}{6}\)
a)\(12:\left\{400:\left[500-\left(125+25×7\right)\right]\right\}\)
\(12:\left\{400:\left[500-300\right]\right\}\)
\(12:2\)
\(6\)
b)\(\left[\left(7-3^3:3^2\right):2^2+99\right]-100\)
\(=\left[4:4+99\right]-100\)
\(=100-100\)
\(=0\)
\(c,3^2×\left[\left(5^2-3\right):11\right]-2^4+2×10^3\)
\(=9×2-16+2×10000\)
\(=18-16+20000\)
\(=20002\)
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Bạn vào đây tham khảo nhé !!!
Dư 0
Ta có:A=\(1+3+3^2+3^3+...+3^{2012}\)
3A=\(3\cdot\left(1+3+3^2+3^3+...+3^{2012}\right)\)
3A=\(3+3^2+3^3+3^4+...+3^{2013}\)
3A-A=\(\left(3+3^2+3^3+3^4+...+3^{2013}\right)-\left(1+3+3^2+3^3+...+3^{2012}\right)\)
2A=\(3+3^2+3^3+3^4+...+3^{2013}-1-3-3^2-3^3-...-3^{2012}\)
2A=\(\left(3-3\right)+\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2012}-3^{2012}\right)+\left(3^{2013}-1\right)\)
2A=\(0+0+0+...+0+3^{2013}-1\)
2A=\(3^{2013}-1\)
A=\(\frac{3^{2013}-1}{2}\)
B=\(3^{2013}\div2\)
B=\(\frac{3^{2013}}{2}\)
VậyB-A=\(\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}\)
\(B-A=\frac{3^{2013}-\left(3^{2013}-1\right)}{2}\)
\(B-A=\frac{3^{2013}-3^{2013}+1}{2}\)
\(B-A=\frac{1}{2}=0,5\)
kq la 1