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Khách

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8 tháng 7 2015

a)A=1/10+1/15+...+1/120

=2(1/20+1/30+...+1/240)

=2(1/4*5+1/5*6+...+1/15*16)

=2*(1/4-1/5+1/5-1/6+...+1/15-1/16)

=2*[(1/4-1/16)+(1/5-1/5)+...+(1/15-1/15)]

=2*[(4/16-1/16)+0+...+0]

=2*3/16=3/8

b) B=1+1/3+1/6+...+1/1225

=2(1/2+1/6+1/12+...+1/2450)

=2(1/1*2+1/2*3+...+1/49*50)

=2*[1-1/2+1/2-1/3+...+1/49-1/50]

=2*[(1-1/50)+(1/2-1/2)+...+(1/49-1/49)]

=2*[(50/50-1/50)+0+...+0]

=2*49/50=49/25

8 tháng 7 2015

a,\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)\)

\(\frac{1}{2}A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\)

\(\frac{1}{2}A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\)

\(\frac{1}{2}A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\)

\(\frac{1}{2}A=\frac{1}{4}-\frac{1}{16}\)\(\frac{1}{2}A=\frac{3}{16}\)suy ra \(A=\frac{3}{16}:\frac{1}{2}=\frac{3}{8}\)

B thì cậu có thể làm nhiều cách 

15 tháng 3 2019

A  em tự tính nhé 

b) B = 1+ 3 + 32+...+399

 3B = 3+ 32+33+...+3100

do đó 3B-B= (3+32+33+...+3100) - ( 1+3+32+...+399)

                  2B= 3100-1

                      B= (3100-1) : 2

c) \(C=1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\)

    \(C=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}\)

   \(C=1+\frac{1}{2}.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)\)

    \(C=1+\frac{1}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)\)

     \(C=1+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

   \(C=1+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{x+1}\right)\)

Phần c thế này thôi vì ko có giá trị x cụ thể .

d) \(D=\frac{9}{8}.\frac{16}{15}.\frac{25}{24}.....\frac{8100}{8099}\)

   \(D=\frac{9.16.25....8100}{8.15.24....8099}\)

\(D=\frac{3.3.4.4.5.5....90.90}{2.4.3.5.4.6.....89.91}\)

\(D=\frac{\left(3.4.5...90\right).\left(3.4.5...90\right)}{\left(2.3.5...89\right).\left(4.5.6...91\right)}\)

\(D=\frac{3.4.5...90}{2.3.4...89}.\frac{3.4.5...90}{4.5.6...91}\)

\(D=\frac{90}{2}.\frac{3}{91}\)

\(D=45.\frac{3}{91}=\frac{135}{91}\)

11 tháng 2 2018

Ta có: \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}...\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}....\frac{1558}{1560}\)

\(=\frac{1.4.2.5....38.41}{2.3.3.4....39.40}=\frac{\left(1.2.3..38\right)\left(4.5...41\right)}{\left(2.3.4...39\right)\left(3...40\right)}=\frac{41}{39.3}=\frac{41}{117}\)

11 tháng 2 2018

 \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}\frac{18}{20}.\frac{28}{30}.........\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...............\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3.4......38\right)\left(4.5.6.7..........41\right)}{\left(2.3.4.5.........39\right)\left(3.4.5.6.........40\right)}\)

\(=\frac{1.41}{39.3}\)

\(=\frac{41}{117}\)

Vậy \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)=\frac{41}{117}\)

31 tháng 3 2019

1) a) A=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)

c) C=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(C=1-\frac{1}{101}\)

\(C=\frac{100}{101}\)

d) Sửa đề: thay \(\frac{3}{92.98}\)=\(\frac{3}{92.95}\)

\(D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}\)

\(D=\frac{1}{2}-\frac{1}{95}\)

\(D=\frac{95-2}{190}=\frac{93}{190}\)

Các bài trên áp dụng theo tính chất: \(\frac{a}{b\left(b+a\right)}\frac{1}{b}-\frac{1}{b+a}\)

13 tháng 5 2019

Ta co:

\(\frac{1}{2}A=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}+\frac{1}{90}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}+\frac{1}{9.10}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{10}=\frac{13}{20}\Rightarrow A=\frac{13}{10}.\)

13 tháng 5 2019

\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{36}+\frac{1}{45}\)

\(A=\frac{2}{4}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{72}+\frac{2}{90}\)

\(A=\frac{2}{2.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{8.9}+\frac{2}{9.10}\)

\(A=2\left(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(A=2.\frac{2}{5}\)

\(A=\frac{4}{5}\)

~ Học tốt ~ K cho mk nhé! Thank you.

8 tháng 5 2016

A=1/2