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a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
a) 11.2+12.3+13.4+....+199.10011.2+12.3+13.4+....+199.100
= 11−12+12−13+13−14+....+199−110011−12+12−13+13−14+....+199−1100
=11+0+0+...+0−110011+0+0+...+0−1100
=1−11001−1100
= 99100
Ta có:
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
=1-\(\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3}\right)-...-\left(\dfrac{1}{99}+\dfrac{1}{99}\right)-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{100}{100}-\dfrac{1}{100}=\dfrac{99}{100}\)
Đặt :
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+................+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}\)
\(A=\dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\)
\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{99.100}< 1\rightarrowđpcm\)
A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)
A=\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)
A=\(\dfrac{1}{1}\)-\(\dfrac{1}{100}\)=\(\dfrac{100}{100}\)-\(\dfrac{1}{100}\)=\(\dfrac{99}{100}\)
Vậy: A=\(\dfrac{99}{100}\)
cho mình hỏi tại sao lại lấy \(\dfrac{1}{1}-\dfrac{1}{100}=\dfrac{100}{100}-\dfrac{1}{100}=\dfrac{99}{100}\)
mà ko lấy các số còn lại như : \(\dfrac{1}{2}-\dfrac{1}{99}...\)
\(a=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Ta có: \(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}\)
\(=\dfrac{a+1-a}{a\left(a+1\right)}\)
\(=\dfrac{1}{a\left(a+1\right)}\) (đpcm)
Ta được:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+...-\dfrac{1}{100}\) \(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
1)Tính
a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
2) tìm x
\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)
\(\dfrac{4}{5}x=0\)
\(x=0:\dfrac{4}{5}\)
\(x=0\)
b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{4}\)
1. Tính:
a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{1}-\dfrac{1}{10}\)
= \(\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
b. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
= \(\dfrac{1}{1}-\dfrac{1}{100}\)
= \(\dfrac{100}{100}-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
2. Tìm x, biết:
a. \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{14}{5}\)
\(x=\dfrac{14}{5}:\dfrac{4}{5}\)
\(x=\dfrac{14}{5}.\dfrac{5}{4}\)
\(x=14.\dfrac{1}{4}\)
\(x=\dfrac{14}{4}\)
Vậy \(x=\dfrac{14}{4}\)
b. \(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}\)
\(\dfrac{2}{5}x=\dfrac{62}{20}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{10}.\dfrac{5}{2}\)
\(x=\dfrac{31}{2}.\dfrac{2}{2}\)
\(x=\dfrac{31}{2}.1\)
\(x=\dfrac{31}{2}\)
Vậy \(x=\dfrac{31}{2}\)
bài này mk tự làm ko sao chép trên mạng
nếu thấy đúng thì tick đúng cho mk nha
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{100-1}{100}\)
\(=\dfrac{99}{100}\)