Gấp ah!
Tìm y
3/4 + 3/28 + 3/70 + 3/130 +...+3/y = 33/34
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
\(\Rightarrow B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\)
\(\Rightarrow B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\)
\(\Rightarrow B=1-\frac{1}{19}=\frac{18}{19}\)
Vậy \(B=\frac{18}{19}\)
Chúc bn học tốt
B = \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\)
= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{13}-\frac{1}{16}\)
= \(1-\frac{1}{16}\)
= \(\frac{15}{16}\)
a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{5}\)
\(x=\dfrac{1}{2}-\dfrac{6}{5}\)
\(x=-\dfrac{7}{10}\)
b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)
\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)
\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)
\(x=\dfrac{60}{13}:\dfrac{12}{13}\)
\(x=5\)
Ta có:
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}\)
\(=\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+\frac{3}{10\times13}+\frac{3}{13\times16}\)
\(=\frac{4-1}{1\times4}+\frac{7-4}{4\times7}+\frac{10-7}{7\times10}+\frac{13-10}{10\times13}+\frac{16-13}{13\times16}\) \(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
Chắc chắn đúng nhé !
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{304}.\)
\(=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{16.19}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{13-10}{10.13}+.....+\frac{19-16}{16.19}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{16}-\frac{1}{19}\)
\(=1-\frac{1}{19}=\frac{18}{19}\)
3/4+3/28+3/70+3/130+...+3/304
= 3 /1.4 + 3/4.7 + 3/7.10 + 3/10.13 +....+ 3 /16.19
= 1 -1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ...+1/16 - 1/19
= 1 - 1/19
=18/19
\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\\ =\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\\ =1-\dfrac{1}{19}=\dfrac{18}{19}\)
\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\)
\(=\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16}+\dfrac{3}{16\times19}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\)
\(=1-\dfrac{1}{19}\)
=\(\dfrac{18}{19}\)
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+...+\frac{3}{10300}\)
\(=\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{100\times103}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}=\frac{102}{103}\)
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{418}+\frac{3}{550}\)
\(\Leftrightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{19.22}+\frac{3}{22.25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
Nhớ k cho m nhé!
Bài này là của lớp 6 nên bạn học lớp 5 có thể bỏ qua nhé
Tìm y:
\(\frac34\) + \(\frac{3}{28}\) + \(\frac{3}{70}\) + \(\frac{3}{130}\) + ... + \(\frac{3}{y}\) = \(\frac{33}{34}\)
\(\frac{3}{1.4}\) + \(\frac{3}{4.7}\) + \(\frac{3}{7.10}\) + \(\frac{3}{10.13}\) + ... + \(\frac{3}{y}\) = \(\frac{33}{34}\)
Xét dãy số: 1; 4; 7; 10;...
mst1 = 1 = 3 - 2
mst2 = 4 = 3.2 - 2
mst3 = 7 = 3.3 - 2
mst4 = 10 = 3.4 - 2
.................................
stn = 3.n - 2
Xét dãy số: 4; 7; 10;...
mst1 = 4 = 3 + 1
mst2 = 7 = 3.2 + 1
...................................
mstn = (3.n + 1)
VT = \(\frac{3}{1.4}\) + \(\frac{3}{4.7}\) + ... + \(\frac{3}{\left(3n-2\right).\left(3n+1\right)}\)
Theo bài ra ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\cdots+\frac{3}{\left(3n-2\right)\left(3n+1\right)}\) = \(\frac{33}{34}\)
\(\frac11-\frac14\) + \(\frac14-\frac17\)+ ... + \(\frac{1}{3n-2}\) - \(\frac{1}{3n+1}\) = \(\frac{33}{34}\)
\(\frac11\) - \(\frac{1}{3n+1}\) = \(\frac{33}{34}\)
\(\frac{1}{3n+1}\) = 1 - \(\frac{33}{34}\)
\(\frac{1}{3n+1}\) = \(\frac{1}{34}\)
3n + 1 = 34
3n = 34 - 1
3n = 33
n = 33: 3
n = 11
y = (3n - 2),(3n + 1)
y = (3.11 - 2).(3.11 + 1)
y = (33 - 2).(33 + 1)
y = 31.34
y = 1054