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\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{20}.\left(1+2+....+20\right)\)
\(=1+\frac{1}{2}\times\frac{2.3}{2}+\frac{1}{3}\times\frac{3.4}{2}+...+\frac{1}{20}\times\frac{20.21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{\left(2+21\right).20:2}{2}=\frac{230}{2}=115\)
Số cuối là
\(\frac{1}{10}.\left(1+2+3+...+10\right)\) hay \(\frac{1}{20}.\left(1+2+3+...+20\right)\) ??
\(5^x=125\)
\(5^x=5^3\)
=> x=3 ( vì cơ số 5>1)
\(3^2.x=81\)
\(9x=81\)
\(x=81:9\)
\(x=9\)
a) 85 . 127 + 5 . 127 . 3
= (85 + 15) . 127
= 100 . 127
= 12700
a) 85 . 127 + 5 . 127 . 3
= (85 + 15) . 127
= 100 . 127
= 12700
b) 1/2 + 5/6 + 11/12 +19/20 + 29/30 + 41/42 + 55/56 + 71/72 + 89/90
1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
1-1/10
9/10
ta gọi biểu thức là A
2A=2^11+2^10+2^9+2^8+...+2+1+0
2A-A=2^11-0
A=2^11=2048
Ta có:\(M=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2M=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\)
\(2M-M=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)
\(M=2-\dfrac{1}{2^9}\)
\(M=\dfrac{1023}{512}\)
Đặt A = \(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
Nhân cả hai với 2 ta được :
2A = \(2\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> 2A = \(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
Trừ 2A cho A , ta được :
2A - A = \(\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}\)
A=1/2+1/2^2+1/2^3+.............+1/2^10
2A=1+1/2+1/2^2+.............+1/2^9
2a-a=(1/2+1/2^2+1/2^3.....+1/2^10)-(1+1/2+1/2^2............+1/2^9)
=>a=1-1/2^10
=>a=1-1/1024
=>A=1023/1024