Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
\(\dfrac{1}{2\cdot3}x+\dfrac{1}{3\cdot4}x+...+\dfrac{1}{49\cdot50}x=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\\ x\cdot\dfrac{12}{25}=1\\ x=1:\dfrac{12}{25}=1\cdot\dfrac{25}{12}=\dfrac{25}{12}\)
Lời giải:
Ta có:
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy ta có đpcm.
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}\)\(=\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)
CM công thức :
\(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)Nhận xét :
\(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{100.101}=\dfrac{1}{100}-\dfrac{1}{101}\)
\(\Rightarrow\)\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(\dfrac{\Rightarrow1}{2}-\dfrac{1}{101}\)
=\(\dfrac{101}{202}-\dfrac{2}{202}=\dfrac{99}{202}\)
~ chúc bn học tốt~
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{299}{600}\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{299}{600}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{300}{600}-\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{1}{600}\)
=> x + 1 = 600
x = 600 - 1
x = 599
Vậy x = 599
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{2}{5}\)
Nhận xét thấy:
\(\dfrac{1}{1.2}\)= 1-\(\dfrac{1}{2}\); \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\);...
Ta có
A= 1-\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
A= 1- \(\dfrac{1}{6}\)
A= \(\dfrac{5}{6}\)
Vậy A= \(\dfrac{5}{6}\)
CAU NAY RAT DE NHA BAN
A=\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)
A=1-\(\dfrac{1}{6}\)
=>A=\(\dfrac{5}{6}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..................+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
Ta có:
\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(=\left(1-\dfrac{1}{99}-1-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\right)\)
\(=\left(-\dfrac{1}{99}-\dfrac{1}{98}\right)\cdot\dfrac{3}{10}=\dfrac{-197\cdot3}{9702\cdot10}=\dfrac{-197}{32340}\)
x-(1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2022.2023)= -2024/2023
x-(1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/2022-1/2023)=-2024/2023
x-(1-1/2023)=-2024/2023
x-2022/2023=-2024/2023
x = -2024/2023+2022/2023
x = -2/2023
Vậy x = -2/2023
:(((