Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b,\frac{10}{99}\)+\(\frac{11}{199}\)+\(\frac{12}{299}\).\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{-1}{6}\)
Ta có: B = \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}\)
=> B = \(\frac{6}{3.5}\)+ \(\frac{6}{5.7}\)+ \(\frac{6}{7.9}\)+ \(\frac{6}{9.11}\)
=>B =\(3.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
=> B = \(3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
=> B = \(3.\left(\frac{1}{3}-\frac{1}{11}\right)\)
=> B = \(3.\frac{8}{33}\)
=> B = \(\frac{8}{11}\)
Vậy: B = \(\frac{8}{11}\)
c) \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\)
\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=2\left(1-\frac{1}{16}\right)\)
\(=2.\frac{15}{16}\)
\(=\frac{15}{8}\)
Vậy A=\(\frac{15}{8}\)
a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
\(B=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+...+\frac{3}{20.22}\)
\(=\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+...+\frac{1}{20.22}\)
\(=\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{20}-\frac{1}{22}\right)\)
\(=\frac{1}{4}-\frac{1}{22}\)
\(=\frac{9}{44}\)
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) \(D=\frac{9}{8}.\frac{16}{15}.\frac{25}{24}.....\frac{8100}{8099}\)
\(D=\frac{9.16.25....8100}{8.15.24....8099}\)
\(D=\frac{3.3.4.4.5.5....90.90}{2.4.3.5.4.6.....89.91}\)
\(D=\frac{\left(3.4.5...90\right).\left(3.4.5...90\right)}{\left(2.3.5...89\right).\left(4.5.6...91\right)}\)
\(D=\frac{3.4.5...90}{2.3.4...89}.\frac{3.4.5...90}{4.5.6...91}\)
\(D=\frac{90}{2}.\frac{3}{91}\)
\(D=45.\frac{3}{91}=\frac{135}{91}\)
bạn phải cho số cuối cùng thì mình mới làm được , nếu không có thì giáo viên của bạn cho sai đề
Ta có
\(\frac{2}{3\cdot4}=\frac{2}{\left(1+2\right)+\left(1+3\right)}\)
\(\frac{2}{4\cdot5}=\frac{2}{\left(2+2\right)\cdot\left(2+3\right)}\)
...
Phân số thứ n là \(\frac{2}{\left(n+2\right)\cdot\left(n+3\right)}\)\(n\in N\)
Phân số thứ 50 là \(\frac{2}{\left(50+2\right)\cdot\left(50+3\right)}=\frac{2}{52\cdot53}\)
\(\Rightarrow\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{52\cdot53}\)
\(=2\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...\frac{1}{52\cdot53}\right)\)
\(=2\cdot\left(\frac{1}{3}-\frac{1}{4}+...+\frac{1}{52}-\frac{1}{53}\right)\)
\(=2\cdot\left(\frac{1}{3}-\frac{1}{53}\right)=\left(\frac{50\cdot2}{159}\right)=\frac{100}{159}\)
1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)
=4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)
= 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)
=4-\(\left(1-\frac{1}{9}\right)\)
= 4-\(\frac{8}{9}\)
= \(\frac{7}{9}\)
a)\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{49}{100}\)
\(=\frac{49}{200}\)
b)\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{201}-\frac{1}{205}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{205}\right)\)
\(=\frac{1}{4}\cdot\frac{204}{205}\)
\(=\frac{51}{205}\)
c)\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3\cdot\frac{32}{99}\)
\(=\frac{32}{33}\)
d)tương tự bạn nhân với 4/3 nhé
a ) 1/2 .4 + 1/4 . 6 + 1/6 . 8 + .........+ 1/98 . 100
= 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ........+ 1/98 - 1/100
= 1/2 - 1/100
= 49/100
b ) 1/1 . 5 + 1/5 . 9 + 1/9 . 13 + ......+ 1/201 . 205
= 1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13+ ..... + 1/201 - 1/205
= 1 - 1/205
= 204/205
c ) 6/3 . 5 + 6/5 . 7 + 6/7 . 9 + ...... + 6/97 . 99
= 6/3 - 6/5 + 6/5 - 6/7 + 6/7 -6/9 + ........ + 6/97 - 6/99
= 6/3 - 6/99
= 64/33
d ) 4/8 . 11 + 4/11 . 14 + 4/14 . 17 + ......... 4/98 . 101
= 4/8 - 4/11 + 4/11 - 4/14 + 4/14 - 4/17 + .......+ 4/98 - 4/101
= 4/8 - 4/101
= 93/202
a) \(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+....+\frac{2}{98\times100}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
= \(\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\frac{98}{200}=\frac{49}{200}\)
a,\(\frac{100}{101}\)
c,\(\frac{27}{20}\)
d,\(\frac{16}{11}\)
,
c, 3/10 + 3/40 + 3/88 + ...+ 3/340
=> 3/10 + 3/44 + 3/88 + .... + 3/340 = 3/2x5 + 3/5x8 + 3/8x11 + .....+ 3/17x 20
= 1/2 - 1/5 + 1/5 - 1.8 + 1/8 - 1/11 +.......+ 1/17 - 1/20
= 1/2 - 1/20 = 9/20
d, 6/15 + 6/35 + 6/63 + 6/99
= 3 ( 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11)
= 3( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
= 3( 1/3 - 1/11)
= 3 x 8/33 = 8/11