Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) A=1+22+24+.................+2100
2A=(1+22+24+.................+2100)
2A=2+23+...+2101
2A-A=(2+23+...+2101)-(1+22+24+.................+2100)
A=2101-1
b)bạn tự làm
c) C=-1/90-1/72-1/50-1/42-1/30-1/20-1/12-1/6-1/2
\(=-\left(\frac{1}{90}+\frac{1}{72}+...+\frac{1}{2}\right)\)
\(=-\left(\frac{1}{10.9}+\frac{1}{9.8}+...+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+...+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{10}-1\right)\)
\(=-\left(-\frac{9}{10}\right)=\frac{9}{10}\)
Bài 2:
cứ tính lần lượt là ra
a) 52x-3 -2.52 = 52.3
52x.125 = 52.3 + 2.52
52x.125 = 125
52x = 1 = 50
=> 2x = 0 => x =0
b) \(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
A = 1/5 - 1/12
A = 7/60
C=1-2+22-23+....+2100
D=1/5-1/52+1/53+.....+1/5101
E=4/2.5+4/5.8+..........+4/302.305
C=1-2+22-23+....+2100
D=1/5-1/52+1/53+.....+1/5101
E=4/2.5+4/5.8+..........+4/302.305
A = 1/5.6 + 1/6.7 + ...... + 1/11.12
= 1/5 - 1/6 + 1/6 - 1/7 + ..... + 1/11 - 1/12
= 1/5 - 1/12
= 7/60
Có : 2B = 1+1/2+1/2^2+.....+1/2^2015
B = 2B - B = (1+1/2+1/2^2+.....+1/2^2015) - (1/2+1/2^2+.....+1/2^2016)
= 1 - 1/2^2016
Tk mk nha
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.20}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
a. \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{17.20}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{17}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
b. \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{4}-\dfrac{1}{10}\)
\(=\dfrac{3}{20}\)
c. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)
\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)
\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=4\left(1-\dfrac{1}{49}\right)\)
\(=4.\dfrac{48}{49}\)
\(=\dfrac{192}{49}\)