Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c) x=-2 nha
d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)
=\(\frac{1}{5}\)-\(\frac{1}{12}\)= \(\frac{7}{60}\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+.....+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}\)
\(A=\frac{7}{60}\)
Ta có:
A = \(\frac{1}{5.6}\)+ \(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)
Bạn xem lời giải của mình nhé:
Giải:
\(A=\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\\ =\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\\ =\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\\ =\frac{1}{5}-\frac{1}{12}=\frac{12-5}{60}=\frac{7}{60}\)
Chúc bạn học tốt!
mình biến đởi phần trong |......| rồi bạn thay vào nha
1/30 + 1/42 + 1/56 + 1/72 +1/ 90 + 1/110 + 1/132
=1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 +1/ 10.11
=1/5 -1/6 +1/6 - 1/7 +......+1/10 - 1/11
=1/5 - 1/11=11/55 - 5/55 =6/ 55
thay vào |....|=> |6/55 - x | = 2/3 => mở ra 2 trường hợp mà tính nha
chúc hok tốt
=>(1/5.6+1/6.7+1/7.8+1/9.10+1/10.11+1/11.12)-x=2/3
=>(1/5-1/+1/6-1/7+...+1/11-1/12)-x=2/3
=>(1/5-1/12)-x=2/3
=>7/60-x=2/3
=>x=7/60-2/3
=>x=-11/20
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
\(A=\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{5}...+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}\)
\(A=\frac{7}{60}\)
A = \(\frac{1}{5.6}+\frac{1}{6.7}+...+\)\(\frac{1}{10.11}+\frac{1}{11.12}\)
A = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\)\(\frac{1}{11}-\frac{1}{12}\)
A = \(\frac{1}{5}-\frac{1}{12}\)
A = \(\frac{7}{60}\)
a) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}\)
\(=\frac{12}{60}+\frac{-5}{60}\)
\(=\frac{7}{60}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{2}{3}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
a) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-...-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
A = 1/5 + [1/5.6 + 1/6.7 + ... + 1/12.13]
A = 1/5 + [1/5-1/6+1/6-1/7+...+1/12-1/13]
A = 1/5 + [1/5-1/13]
A = 1/5 + 8/65
A = 21/65
1/90 đó bn
Nếu đúng thì tk nha !!!!!
\(\frac{1}{90}hihihi\)là số thích hợp nhất.
Ta có:\(\frac{1}{2},\frac{1}{12},\frac{1}{30},\frac{1}{56},\frac{1}{90},\frac{1}{132}\)
Vậy dấu ? là phân số\(\frac{1}{90}\)
~Chúc bạn học giỏi~