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=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+............+1/9+1/10
=1-1/10
=10/10-1/10
=9/10
Bài làm:
\(\frac{1}{1\times2}+\frac{1}{2\times3}\)\(+\frac{1}{3\times4}+\frac{1}{4\times5}\)\(+...\frac{1}{9\times10}\)
\(=\frac{1}{1}-\frac{1}{2}\)\(+\frac{1}{2}-\frac{1}{3}\)\(+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}\)\(-\frac{1}{5}\)\(+...\frac{1}{9}-\frac{1}{10}\)
\(=\)\(\frac{1}{1}-\frac{1}{10}\)
\(=\frac{9}{10}\)
1/1 x 2 + 1/2 x 3 + 1/3 x 4 + .... + 1/9 x 10
= 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 + ... + 1/9 - 1/10
= 1 - 1/10
= 9/10
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{9}{10}\)
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{100\cdot101}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
Ta đặt biểu thức là:
A = 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + .... + 1/9 x 10
A = 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 + ... + 1/9 - 1/10
A = 1 - 1/10
A = 9/10
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{15.16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{8\times9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{1}-\frac{1}{9}=\frac{8}{9}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{1}-\frac{1}{9}\)
\(=\frac{8}{9}\)
1/1.2+1/2.3+1/3.4+...+1/999.1000+1
=1-1/2+1/2-1/3+1/3-1/4+...+1/998-1/999+1/999-1/1000+1
=1-1/1000+1
=999/1000+1
=1999/1000
Chuẩn ko cần chỉnh
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{999\times1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{999}{1000}+1\)
\(=\frac{1999}{1000}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2018.2019}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
Dấu \(.\)là dấu nhân .
Ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2019}{2019}-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
~ Ủng hộ nhé