K
Khách
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7 tháng 8 2015
a) \(\frac{13}{7}-\frac{1}{2}\times\frac{13}{7}+\frac{3}{2}\times\frac{13}{7}\)
\(=\frac{13}{7}\times\left(1-\frac{1}{2}+\frac{3}{2}\right)\)
\(=\frac{13}{7}\times2\)
\(=\frac{26}{7}\)
b) \(\frac{1}{15}\times\left(\frac{3}{7}+\frac{5}{19}\right)+\frac{3}{7}\times\left(\frac{5}{19}-\frac{1}{15}\right)\)
\(=\frac{1}{15}\times\frac{3}{7}+\frac{1}{15}\times\frac{5}{19}+\frac{3}{7}\times\frac{5}{19}-\frac{3}{7}\times\frac{1}{15}\)
\(=\frac{5}{19}\times\left(\frac{1}{15}+\frac{3}{7}\right)\)
\(=\frac{5}{19}\times\frac{52}{105}\)
\(=\frac{52}{399}\)
c) \(\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}\)
\(=5\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\right)\)
\(=5\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5\times\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=5\times\frac{49}{100}\)
\(=\frac{49}{20}\)
Lần sau nên đăng ít thôi
C
0
CM
31 tháng 10 2018
3/2+7/6+13/12+21/20+...+133/132
=1+ 1/2 + 1+ 1/6 + 1+ 1/12 + 1+ 1/20+......+ 1+ 1/132
=(1+ 1 + 1+.....+ 1) + ( 1/2+ 1/6+1/12+ 1/20+....+ 1/132)
= 11+ ( 1/1×2+ 1/2×3+1/3×4+1/4×5+.....+1/11×12)
=11+ ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+......+1/11-1/12)
=11+(1-1/12)
= 11+11/12
=143/12
CM
8 tháng 1 2018
3/2+7/6+13/12+21/20+...+133/132 =1+ 1/2 + 1+ 1/6 + 1+ 1/12 + 1+ 1/20+......+ 1+ 1/132 =(1+ 1 + 1+.....+ 1) + ( 1/2+ 1/6+1/12+ 1/20+....+ 1/132) = 11+ ( 1/1×2+ 1/2×3+1/3×4+1/4×5+.....+1/11×12) =11+ ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+......+1/11-1/12) =11+(1-1/12) = 11+11/12 =143/12
DD
4
23 tháng 8 2015
A = 1 + 1/110 + 1 + 1/90 + ... + 1 + 1 /2
A = 10 + 1/1.2+ 1 /2.3 + ... + 1/9.10 + 1/10.11
A = 10 + 1/1 - 1/2 + 1 /2 - 1/3 + ... + 1/9 - 1/10 + 1/10 - 1/11
A = 10 + 1/1 - 1/11
A = 10 + 10/11
A = 120/11
HT
23 tháng 8 2015
A = \(\frac{111}{110}+\frac{91}{90}+\frac{73}{72}+...+\frac{13}{12}+\frac{7}{6}+\frac{3}{2}\)
A = \(\left(\frac{1}{2}+1\right)+\left(\frac{1}{6}+1\right)+\left(\frac{1}{12}+1\right)+....+\left(\frac{1}{110}+1\right)\)
A = (1 + 1 + 1 +...+ 1) + \(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
A = 10 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
A = \(10+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
A = \(10+\left(1-\frac{1}{11}\right)\)
A = \(10+\frac{10}{11}\)
A = \(\frac{120}{11}\)
20 tháng 9 2015
\(\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
= 1 + 2 + 2
= 5
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
KS
1
VT
20 tháng 8 2017
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
Ta có:
\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{9901}{9900}\)
\(A=1+\frac{1}{2}+1+\frac{1}{6}+1+\frac{1}{12}+...+1+\frac{1}{9900}\)\(A=1+1+1+...+1(51c/s)+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(A=51+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=51+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=51+1-\frac{1}{100}\)
\(A=52-\frac{1}{100}\)
\(A=\frac{5199}{100}\)
Cái đoạn 1+1+1+...+1 ( 51 c/s) số tớ ko thể giải thích trên máy tính đc nên bn tự suy nghĩ nhé:)))