Tinh A=\(\frac{10}{3.8}\)+\(\frac{10}{8.13}\)+\(\frac{10}{13.18}\)+\(\frac{10}{18.23}\)+\(\frac{10}{23.28}\)
\(B=\)\(\frac{10}{3.8}\)\(+\frac{10}{8.13}\)\(+\frac{10}{13.18}\)\(+\frac{10}{18.23}\)\(+\frac{10}{23.28}\)
\(B=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(B=2\left[\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{28}\right]=\frac{25}{42}\)
B = 10/3.8 + 10/8.13 + 10/13.18 + 10/18.23 + 10/23.28
= 2.( 5/3.8 + 5/8.13 + 5/13.18 + 5/18.23 + 10/23.28 )
= 2.( 1/3 -1/8 + 1/8 - 1/13 + 1/13 - 1/18 + 1/18 - 1/23 + 1/23 - 1/28 )
= 2.( 1/3 - 1/28 )
= 2. 25/84
= 25/42
\(\frac{24}{42}\) nha bn chúc hc tốtttt
tìm x
\(\frac{5x}{3.8}+\frac{5x}{8.13}+\frac{5x}{13.18}+\frac{5x}{18.23}+\frac{5x}{23.28}+\frac{5x}{28.33}=\frac{-7}{6}\)
\(\Rightarrow x\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...........+\frac{1}{28}-\frac{1}{33}\right)=\frac{-7}{6}\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{-7}{6}\)
\(\Rightarrow x.\frac{10}{33}=\frac{-7}{6}\)
\(\Rightarrow x=\frac{-7}{6}:\frac{10}{33}\)
\(\Rightarrow x=\frac{-231}{60}\)
\(\frac{10}{3.8}\)x\(\frac{10}{8.13}\)x\(\frac{10}{13.18}+...+\frac{10}{48.53}\)
Tinh tong
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)
\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=2.\frac{50}{159}=\frac{100}{159}\)
Bài 1:
Cho A= 2015/2016+2016/2017+2017/2018+2018/2019
Chứng minh A >4
Bài 2:
Tính: A=10/3.8+10/8.3+10/13.18+10/18.23+10/23.28
Bài 3:
Tính các số nguyên n để phân số n+6/n+1 là số nguyên.
Các bạn có thể làm 1 bài cũng được.
Bài 3
\(\frac{n+6}{n+1}=\frac{n+1+5}{n+1}=\frac{n+1}{n+1}+\frac{5}{n+1}\)
\(=1+\frac{5}{n+1}\)
Vậy để \(\frac{n+6}{n+1}\in Z\Rightarrow1+\frac{5}{n+1}\in Z\)
Hay \(\frac{5}{n+1}\in Z\)\(\Rightarrow n+1\inƯ_5\)
\(Ư_5=\left\{1;-1;5;-5\right\}\)
* \(n+1=1\Rightarrow n=0\)
* \(n+1=-1\Rightarrow n=-2\)
* \(n+1=5\Rightarrow n=4\)
* \(n+1=-5\Rightarrow n=-6\)
Vậy \(n\in\left\{0;-2;4;-6\right\}\)
Bài 2:
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\\ =2\left(\frac{1}{3}-\frac{1}{28}\right)\\ =2.\frac{56}{84}\\ =\frac{56}{42}=\frac{28}{21}\)
Bài 5: (1 điểm ) Tính
\(\left(\frac{5^3}{8.13}+\frac{5^3}{13.18}+\frac{5^3}{18.23}+...+\frac{5^3}{93.98}\right).3\frac{17}{125}\)
\(=5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right).\frac{392}{17}\)
\(=5^2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\left(\frac{1}{8}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\times\frac{45}{392}\times\frac{392}{17}\)
\(=25\times\frac{45}{17}\)
\(=\frac{1125}{17}\)
\(5^2.\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{93.98}\right).\frac{392}{5^2}\)
\(\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right).392=\left(\frac{1}{8}-\frac{1}{98}\right).392=45\)
Tính:A=1/7.12+1/12.17+1/17.22+...+1/52.57
B=10/8.13+10/13.18+10/18.23+...+10/253.258
\(A=\frac{1}{7\cdot12}+\frac{1}{12\cdot17}+\frac{1}{17\cdot22}+...+\frac{1}{52\cdot57}\)
\(A=\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+...+\frac{5}{52\cdot57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{52}-\frac{1}{57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{57}\right)=\frac{1}{5}\cdot\frac{50}{399}=\frac{10}{399}\)
\(B=\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+...+\frac{10}{253\cdot258}\)
\(B=\frac{10}{5}\left(\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+...+\frac{5}{253\cdot258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{258}\right)=2\cdot\frac{125}{1032}=\frac{125}{516}\)
*Cái đây giải thích hơi bị " khó hiểu " :
Chỗ mẫu (12 - 7) = (17 - 12) = ... = (57 - 52) = 5
Tử là 1 , mẫu là 5 nên tử/mẫu = 1/5
Hay \(\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+...+\frac{5}{52\cdot57}\right)\)
Còn bạn Trương Bùi Linh thì :
Mẫu = (13 - 8) = (18 - 13) = (23 - 18) = ... = 5
Tử là 10,mẫu là 5 => tử / mẫu = 10/5 = 2
Tính nhanh:
a) \(A=\frac{3}{5.8}\)+ \(\frac{3}{8.11}\)+ \(\frac{3}{11.14}\)+ ... + \(\frac{3}{2006.2009}\)
b) \(B=\frac{1}{6.10}\)+ \(\frac{1}{10.14}\)+ \(\frac{1}{14.18}\)+...+ \(\frac{1}{402.406}\)
c) \(C=\frac{10}{7.12}\)+\(\frac{10}{12.17}\)+\(\frac{10}{17.22}\)+...+\(\frac{10}{502.507}\)
d) \(D=\frac{4}{8.13}\)+\(\frac{4}{13.18}\)+\(\frac{4}{18.23}\)+...+\(\frac{4}{253.258}\)
Nếu ai có giải dùm mình thì giải từng phần nhưng đừng chỉ ghi kết quả nhé~
a,\(\frac{2004}{10045}\)
b,\(\frac{25}{609}\)
c,\(\frac{1000}{3549}\)
d,\(\frac{25}{258}\)
A) \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+.....+\frac{1}{2006}-\frac{1}{2009}=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)các câu sau không theo quy luật bạn phải quy đồng mình đã giải xong câu a) rồi đấy bạn hãy vận dụng nó để giải các câu còn lại nên nhớ câu a) theo quy luật còn các câu còn lại không theo quy luật nên ta quy đồng không phải là mình không biết làm đâu => tại mình mỏi tay quá
Tính :
\(\frac{4}{8.13}\) + \(\frac{4}{13.18}\) + \(\frac{4}{18.23}\) + ... + \(\frac{4}{253.258}\) .
Cứu mình đi! Gấp lắm rồi
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.24}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\frac{125}{1032}\)
\(=\frac{25}{258}\)
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\frac{125}{1032}=\frac{25}{258}\)
a)\(\frac{45^{10}.5^{10}}{75^{10}}\)
b)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\)
c)\(\frac{2^{15}.9^4}{6^3.8^3}\)
d)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)