\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+....+\(\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{1}{3}\)-\(\dfrac{x+1}{324}\)
giúp mình vói ạ
tìm n thuộc N biết:
\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+....+\(\dfrac{1}{n\left(n+1\right)}\)=\(\dfrac{3}{10}\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{n+1}=\dfrac{1}{30}\)
\(\Rightarrow n+1=30\)
\(\Rightarrow n=29\)
Vậy n = 29.
\(\dfrac{1}{x}\) - \(\dfrac{2}{x+1}\) = \(\dfrac{3}{x^2+x}\)
\(\dfrac{1}{x2-3}\) - \(\dfrac{3}{x\left(2x-3\right)}\) = \(\dfrac{5}{x}\)
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
GIúp mình với ạ
a: ĐKXĐ: x<>0; x<>-1
PT =>x+1-2x=3
=>1-x=3
=>x=-2(nhận)
b: Sửa đề: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)
=>x-3=5(2x-3)
=>10x-15=x-3
=>9x=12
=>x=4/3(nhận)
c: ĐKXĐ: x<>0; x<>2
PT =>x(x+2)-x+2=2
=>x^2+2x-x=0
=>x(x+1)=0
=>x=-1
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{1}{3}\) .-.
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{3-1}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{200.201}\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{200.201}\)
=\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{200}-\dfrac{1}{201}\)
=\(\dfrac{1}{3}-\dfrac{1}{201}\)
=\(\dfrac{67}{201}-\dfrac{1}{201}\)
=\(\dfrac{66}{201}\)
---Học Tốt Nha---
Giải:
1/3.4+1/4.5+1/5.6+...+1/200.201
=1/3-1/4+1/4-1/5+1/5-1/6+...+1/200-1/201
=1/3-1/201
=22/67
Chúc bạn học tốt!
BT2: Tìm x\(\in\) N*, biết
2) \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{x.\left(x+1\right)}=\dfrac{299}{600}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{299}{600}\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{299}{600}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{300}{600}-\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{1}{600}\)
=> x + 1 = 600
x = 600 - 1
x = 599
Vậy x = 599
Tính B= \(\dfrac{1}{3.4}\)-\(\dfrac{1}{4.5}\)-\(\dfrac{1}{5.6}\)-\(\dfrac{1}{6.7}\)-\(\dfrac{1}{7.8}\)-\(\dfrac{1}{8.9}\)-\(\dfrac{1}{9.10}\)
B = \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\) - \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\)
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{10}\))
B = \(\dfrac{1}{12}\) - \(\dfrac{3}{20}\)
B = - \(\dfrac{1}{15}\)
Tính tổng:
\(M=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\)
\(M=\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(M=1-\dfrac{1}{7}\)
\(M=\dfrac{6}{7}\)
tham khảo
https://hoc24.vn/cau-hoi/123134145156167.5003535458609#:~:text=l%C3%BAc%2021%3A02-,1,14,-12.3%2B13.4%2B14.5
vào đi
refer
https://hoc24.vn/cau-hoi/123134145156167.5003535458609#:~:text=l%C3%BAc%2021%3A02-,1,14,-12.3%2B13.4%2B14.5
a) tìm n thuộc Z để phân số sau đây là số nguyên\(\dfrac{3}{n-2}\)
b)tìm số y nguyên dương biết:\(\dfrac{3}{y}< \dfrac{y}{7}< \dfrac{4}{y}\)
c)\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{29.30}\)
d)\(\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right).\left(1-\dfrac{1}{6}\right)......\left(1-\dfrac{1}{29}\right).\left(1-\dfrac{1}{30}\right)\)
a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3
\(\Rightarrow\) n - 2 \(\in\) Ư(3)
\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}
n \(\in\){5; -1; 3; 2}
c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
d)\(\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right).\left(1-\dfrac{1}{6}\right).......\left(1-\dfrac{1}{29}\right).\left(1-\dfrac{1}{30}\right)\)\(=\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}....\dfrac{28}{29}\)
\(=\dfrac{3.4.5...28}{4.5.6...29}\)
\(=\dfrac{3}{29}\)