\(\frac{3}{9\cdot15}+\frac{3}{15\cdot21}+\frac{3}{21\cdot27}+.....+\frac{3}{99\cdot105}\)
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25 tháng 7 2019
Phần 1)Đầu tiên bạn nhân B với 1 phần 4 rồi tính đến đoạn gần cuối sẽ ra 1/3 - 1/35 rồi quy đòng rồi tính sẽ ra kêt quả cuối là 32/105 nha
Mình lười lắm nên chỉ help 1 phần thui nha sr
8 tháng 5 2015
A=2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
A= 2 - 1/3 + 1/3 - 1/5 + 1/5 - ... + 2/99 - 2/101
A = 2 - 2/101 = 200/101
B = 3-1/3+1/3-1/5+1/5-...+3/49-3/51
B = 3-3/51(tự tính nhé)
C = 5(5/1.6+5/6.11+5/11.16+....+5/26-5/31
C = 5(5-1/31)(tự tính)
D rút gon cho 2 rồi 3D , sau đó 5(3/.... tương tự các cách làm trên)
2E nhân lên rồi giải giống trên
3F Rồi nhân 4/77 và rút gọn thì tính được
XT
16 tháng 7 2015
a, A= \(\frac{1}{1}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+......+\(\frac{1}{99}\)-\(\frac{1}{100}\)
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+(-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....-\(\frac{1}{99}\)+\(\frac{1}{99}\))
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+0
A=1-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)
23 tháng 2 2018
=> 1/11 - 1/13 + 1/13 - 1/15 + ..... + 1/19 - 1/21 - x + 4 + 221/231 = 7/3
=> 1/11 - 1/21 - x + 4 + 221/231 = 7/3
=> 2099/420 - x = 7/3
=> x = 2099/420 - 7/3 = 373/140
Tk mk nha
20 tháng 5 2020
Bài làm
\(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow2\left(\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{19.21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow2\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow2\left(\frac{1}{11}-\frac{1}{21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow2.\frac{10}{231}-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow\frac{20}{231}-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow\frac{20}{231}-x+\frac{924}{231}+\frac{221}{231}=\frac{539}{231}\)
\(\Leftrightarrow\frac{20}{231}-x+\frac{924}{231}=\frac{539}{231}-\frac{221}{231}\)
\(\Leftrightarrow\frac{20}{231}-x+\frac{924}{231}=\frac{318}{231}\)
\(\Leftrightarrow\frac{20}{231}-x=\frac{318}{231}-\frac{924}{231}\)
\(\Leftrightarrow\frac{20}{231}-x=-\frac{606}{231}\)
\(\Leftrightarrow x=\frac{20}{231}-\frac{606}{231}\)
\(\Leftrightarrow x=-\frac{586}{231}\)
Vậy \(\Leftrightarrow=-\frac{586}{231}\)
2 tháng 5 2019
\(B=\frac{5}{18\cdot21}+\frac{5}{21\cdot24}+\frac{5}{24\cdot27}+...+\frac{5}{123\cdot126}\\ B=\frac{5}{3}\cdot\left(\frac{3}{18\cdot21}+\frac{3}{21\cdot24}+\frac{3}{24\cdot27}+...+\frac{3}{123\cdot126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+\frac{1}{24}-\frac{1}{27}+...+\frac{1}{123}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{1}{18}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{7}{126}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\frac{1}{21}\\ B=\frac{5}{63}\)
1 tháng 7 2018
Sửa đề : \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=18\)
\(\Leftrightarrow\)\(x=18-1\)
\(\Leftrightarrow\)\(x=17\)
Vậy \(x=17\)
Chúc bạn học tốt ~
1 tháng 7 2018
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10.\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+\frac{40}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x=\frac{3}{11}-\frac{40}{55}\)
\(\Leftrightarrow\)\(x=\frac{-5}{11}\)
Vậy \(x=\frac{-5}{11}\)
Chúc bạn học tốt ~
29 tháng 6 2017
a) \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)
\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{19}-\frac{1}{19}+\frac{1}{23}-\frac{1}{23}+\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)
\(=\frac{1}{2}-\frac{2}{27}\)
\(=\frac{23}{54}\)
b) \(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+\frac{1}{20}-...-\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\frac{19}{100}\)
\(=1-\frac{19}{500}\)
\(=\frac{481}{500}\)
2 tháng 3 2016
(2/143+2/195+2/255+2/323+2/399).462-y=19
=> 10/231.462-y=19
=> 20-y=19
=> y= 20-19=1
DUYỆT NHA "T I C K"
=\(\frac{1}{2}\cdot\left(\frac{6}{9\cdot15}+\frac{6}{15\cdot21}+\frac{6}{21\cdot27}+...+\frac{6}{99\cdot105}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{1}{9}-\frac{1}{15}+\frac{1}{15}-\frac{1}{21}+\frac{1}{21}-\frac{1}{27}+...+\frac{1}{99}-\frac{1}{105}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{1}{9}-\frac{1}{105}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{35}{315}-\frac{3}{315}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{32}{315}\right)\)
=\(\frac{1\cdot16}{1\cdot315}\)
=\(\frac{16}{315}\)