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\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
= \(1+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
= \(1+\frac{3}{4}\)
= \(\frac{4}{4}+\frac{3}{4}\)
= \(\frac{7}{4}\)
HỌC TỐT
\(A=1+5+5^2+...+5^{201}\)
\(5A=5+5^2+5^3+...+5^{201}+5^{202}\)
\(4A=5A-A=5^{202}-1\)
\(A=\frac{5^{202}-1}{4}\)
Tính tổng:a)3+3/5+3/25+3/125+3/625
b)M=4/3.7+4/7.11+4/11.15+...+8/95.99
c)N=1/2+1/6+1/12+1/20+...+1/90
Ta có : \(M=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
bài 1
a) 155 - 10.(x+1) = 55
=>10 .(x+1) = 100
=>x + 1 = 10
=>x = 9
còn lại tương tự
25.84=25.(23)4=25.212=217
256.1253=(52)6.(53)3=512.59=517
6255:257=(54)5:(52)7=520:214=56
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)
\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)
\(=2-\frac{1}{5}\)< 2
Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2
\(S=\dfrac{625}{625}+\dfrac{125}{625}+\dfrac{25}{625}+\dfrac{5}{625}+\dfrac{1}{625}\)
\(=\dfrac{781}{625}\)
S = 1 + \(\dfrac{1}{5}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{125}\) + \(\dfrac{1}{625}\)
5.S = 5 +1 + \(\dfrac{1}{5}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{125}\)
5S - S = 5 - \(\dfrac{1}{625}\)
S = ( 5 - \(\dfrac{1}{625}\)) : 4
S = \(\dfrac{781}{625}\)