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a) \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{14}{30}=\frac{7}{15}\)
a)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=2\left(1-\frac{1}{15}\right)\)
\(=2.\frac{14}{15}\)
\(=\frac{28}{15}\)
b)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}+\frac{2}{10.11}+\frac{2}{11.12}\)
\(...\)
\(=1+\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+\frac{1}{5.3}+\frac{1}{3.7}+\frac{1}{7.4}+\frac{1}{4.9}+\frac{1}{9.5}\)
\(=1+1-\frac{1}{5}\)
\(=\frac{10}{5}-\frac{1}{5}\)
\(=\frac{9}{5}\)
Ai thấy đúng thì
phaỉ giải chi tiết chứ nói như nguyentuantai thì bấm áy tính cũng ra thôi!
\(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(=2\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=2\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(=2\left(\left[\frac{1}{3}-\frac{1}{4}\right]+\left[\frac{1}{4}-\frac{1}{5}\right]+\left[\frac{1}{5}-\frac{1}{6}\right]+\left[\frac{1}{6}-\frac{1}{7}\right]+\left[\frac{1}{7}-\frac{1}{8}\right]+\left[\frac{1}{8}-\frac{1}{9}\right]+\left[\frac{1}{9}-\frac{1}{10}\right]\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{10}\right)\)
\(=2\cdot\frac{7}{30}\)
\(=\frac{7}{15}\)
21)
\(\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{8}\right).\left(1+\dfrac{1}{15}\right).....\left(1+\dfrac{1}{9999}\right)\\ =\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{10000}{9999}\\ =\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{100.100}{99.101}\\ =\dfrac{2.3.4.....100}{1.2.3.....99}.\dfrac{2.3.4.....100}{3.4.5.....101}\\ =100.\dfrac{2}{101}\\ =\dfrac{200}{101}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}A=1-\frac{1}{10}\)
\(\frac{1}{2}A=\frac{9}{10}\)
\(A=\frac{9}{10}:\frac{1}{2}\)
\(A=\frac{18}{10}=\frac{9}{5}\)
Ta co:
\(\frac{1}{2}A=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}+\frac{1}{90}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{10}=\frac{13}{20}\Rightarrow A=\frac{13}{10}.\)
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{36}+\frac{1}{45}\)
\(A=\frac{2}{4}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{72}+\frac{2}{90}\)
\(A=\frac{2}{2.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{8.9}+\frac{2}{9.10}\)
\(A=2\left(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=2.\frac{2}{5}\)
\(A=\frac{4}{5}\)
~ Học tốt ~ K cho mk nhé! Thank you.
1/2D=1/2(1/6+1/10+......+1/45)
1/2D=1/12+1/20+1/30+.....+1/90
1/2D=1/3.4+1/4.5+1/5.6+......+1/9.10
1/2D=1/3-1/4+1/4-1/5+1/5-1/6+....+1/9-1/10
1/2D=1/3-1/10
1/2D=7/30
D=7/30:1/2
D=7/15
Ta có:\(D=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(=\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}\)
\(=2.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{10}\right)=2.\frac{7}{30}=\frac{7}{15}\)
Vậy \(D=\frac{7}{15}\)
Đặt A = 1/2 + 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/36 + 1/45
=> 1/2A = 1/4 + 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/72 + 1/90
= 1/4 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/8.9 + 1/9.10
= 1/4 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10
= 1/4 + 1/2 - 1/10
= 5/20 + 10/20 - 2/20
= 13/20
=> A = 13/20 : 1/2 = 13/10
Đặt A = 1/2 + 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/36 + 1/45
=> 1/2A = 1/4 + 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/72 + 1/90
= 1/4 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/8.9 + 1/9.10
= 1/4 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10
= 1/4 + 1/2 - 1/10
= 5/20 + 10/20 - 2/20 = 13/20
=> A = 13/20 : 1/2 = 13/10
A=1+(1/6+1/12+1/20+...+1/90):2
A=1+(1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10):2
A=1+(1/2-1/10):2
A=1+2/5:2
A=1+1/5
A=6/5
Vậy A=6/5 nha bạn
Đúng 100%
k mk nha
Mk nhanh nhất
thừa 1 số 1/45 nha !!