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Tính tổng : 1 + 3 + 5 + 7 + ... + 99.
Bài làm :
= ( 1 + 99 ) + ( 97 + 3 ) + ( 95 + 5 ) + ( 93 + 7 ) + ( 91 + 9 ) + ( 89 + 11 ) + ... , tổng cộng có 25 cặp có tổng là 100.
= 25 x 100.
= 2500.
Số số hạng có trong B là:
(99 - 1) : 2 + 1 = 50 (số)
B = (1 + 99) x 50 : 2
B = 100 x 50 : 2
B = 5000 : 2
B = 2500
1.
a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)
b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)
3.
a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)
b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)
c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)
a) 16 24 − 1 3 = 16 24 − 8 24 = 24 16 − 3 1 = 24 16 − 24 8 = 8 24 = 1 3 24 8 = 3 1 b) 4 5 − 12 60 = 48 60 − 12 60 = 36 60 = 9 15 5 4 − 60 12 = 60 48 − 60 12 = 60 36 = 15 9 3. a) 17 6 − 2 6 = 17 − 2 6 = 15 6 6 17 − 6 2 = 6 17−2 = 6 15 b) 16 15 − 11 15 = 16 − 11 15 = 5 15 = 1 3 15 16 − 15 11 = 15 16−11 = 15 5 = 3 1 c) 19 12 − 13 12 = 19 − 13 12 = 6 12 = 1 2 12 19 − 12 13 = 12 19−13 = 12 6 = 2 1
a, \(\frac{2}{3}+\frac{2}{3}+\frac{6}{3}=\frac{10}{3}\)
b,\(\frac{3}{4}+\frac{3}{4}+\frac{3}{2}=\frac{6}{4}+\frac{3}{2}=\frac{3}{2}+\frac{3}{2}=\frac{6}{2}=3\)
b) 1/3+1/3^2+1/3^3+1/3^4+1/3^5 (goi tong bang M)
3M=1+1/3+1/3^2+1/3^3+1/3^4
3M-M=1-1/3^5
2M=242/243
M=242/243*1/2=121/243
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{77}{60}\)
\(B=\frac{2}{5}X\frac{15}{4}:\frac{5}{16}+\frac{1}{2}=\frac{53}{10}\)
TK NHA
Ai tk mình đi mình bị âm nè mình hứa sẽ k lại!
Cảm ơn trc nha
Ta có:
\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)
Ta lại có:
\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)
\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)
\(=\frac{1.2.3...36}{1.2.3...36}=1\)
Từ đây ta suy ra được
\(A-B=1-1=0\)
( 1- \(\dfrac{1}{15}\))x ( 1 - \(\dfrac{1}{16}\))x(1- \(\dfrac{1}{17}\)) x.....x(1- \(\dfrac{1}{100}\))
= \(\dfrac{14}{15}\) x \(\dfrac{15}{16}\)x \(\dfrac{16}{17}\)x......x\(\dfrac{99}{100}\)
= \(\dfrac{15\times16\times17\times.....\times99}{15\times16\times17\times.....\times99}\) x \(\dfrac{14}{100}\)
= 1 x \(\dfrac{14}{100}\)
= \(\dfrac{7}{50}\)