Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\)\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{51.56}\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{51.56}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{51}-\frac{1}{56}\)
\(5A=1-\frac{1}{56}=\frac{55}{56}\)
\(A=\frac{55}{56}\div5=\frac{55}{56}.\frac{1}{5}=\frac{11}{56}\)
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{46.51}\)
\(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{46.51}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{51}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{51}\right)\)
\(=\frac{1}{5}.\left(\frac{51}{51}-\frac{1}{51}\right)\)
\(=\frac{1}{5}.\frac{50}{51}\)
\(=\frac{10}{51}\)
Chúc bạn học tốt !!!
\(A=\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{496}-\frac{1}{501}\right):5\)
\(A=\left(1-\frac{1}{501}\right):5\)
\(A=\frac{500}{501}:5=\frac{100}{501}\)
Ta có : \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.501}\)
\(\Rightarrow\) \(A=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{496}-\frac{1}{501}\right) \)
\(\Rightarrow\) \(A=\frac{1}{5}\left(1-\frac{1}{501}\right)\)
\(\Rightarrow\) \(A=\frac{1}{5}.\frac{501-1}{501}=\frac{1}{5}.\frac{500}{501}\)
\(\Rightarrow\) \(A=\frac{1.500}{5.501}=\frac{20}{1.501}=\frac{20}{501}\)
Vậy \(A=\frac{20}{501}\)
Nhân A với 5 ta được
A .5=5.(1/1.6 +1/6.11+1/11.16+....+1/496.561)
A .5=5/1.6+5/6.11+5/11.16+...+5/496.561
A .5=1-1/6+1/6-1/11+1/11-1/16+...+1/496-1/561
A .5=1-1/561
A .5=560/561
A =560/561 : 5 =112/561
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.561}\)
\(=\left(\frac{1}{1}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{16}\right)+...+\left(\frac{1}{496}-\frac{1}{561}\right)\)
\(=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{496}-\frac{1}{561}\)
\(=\frac{1}{1}-\frac{1}{561}\)
\(=\frac{561}{561}-\frac{1}{561}\)
\(=\frac{560}{561}\)
\(\Rightarrow\) Vậy, \(A=\frac{560}{561}\)
Ta có :
\(\frac{5}{1.6}+\frac{5}{6.11}+................+\frac{5}{\left(5.x+1\right).\left(5.x+6\right)}=\)\(\frac{50}{41}\)
=> \(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...............+\frac{1}{5.x+1}-\frac{1}{5.x+6}\) = \(\frac{50}{41}\)
=> \(1-\frac{1}{5.x+6}=\frac{50}{41}\)
=> \(\frac{1}{5.x+6}=\frac{-9}{41}\)................ mình ko tìm ra vì p/s kia ko có tử là 1
bạn xem lại đề bài giúp mình nha
1-1/6+1/6-1/11+...+1/5x+1-1/5x+6=2005/2006
1-1/5x+6=1-1/2006
5x+6=2006
5x=2000
x=400
\(1-\frac{1}{5x+6}=\frac{2005}{2006}\Leftrightarrow5x+6=2006\Leftrightarrow x=400\)
\(\frac{5x}{1.6}+\frac{5x}{6.11}+\frac{5x}{11.16}+\frac{5x}{16.21}=\frac{1}{25}\)
\(x\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{21}\right)=\frac{1}{25}\)
\(\frac{20}{21}x=\frac{1}{25}\)
\(x=\frac{1}{25}:\frac{20}{21}=.....\)
Ta có:
Số hạng thứ nhất: \(\frac{1}{1.6}=\frac{1}{\left(5.1-4\right).\left(5.1+1\right)}\)
Số hạng thứ 2: \(\frac{1}{6.11}=\frac{1}{\left(5.2-4\right).\left(5.2+1\right)}\)
Số hạng thứ 3: \(\frac{1}{11.16}=\frac{1}{\left(5.3-4\right)+\left(5.3+1\right)}\)
.......
Số hạng thứ n = \(\frac{1}{\left(5.n-4\right)+\left(5.n+1\right)}\)
Vậy số hạng 100 của dãy đó là: \(\frac{1}{\left(5.100-4\right).\left(5.100+1\right)}=\frac{1}{496.501}\)