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\(Đề:\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{99.100}\)
\(= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . \ldots + \frac{1}{99} - \frac{1}{100}\)
\(= 1 - \frac{1}{100}\)
\(= \frac{99}{100}\)
A = \(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+... + \(\frac{1}{99}\)-\(\frac{1}{100}\)
A = \(\frac{1}{1}\)-\(\frac{1}{100}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm đây
A = \(\frac{99}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}=\frac{99}{100}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
A = \(1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....++\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}=\frac{9}{10}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(C=1-\frac{1}{100}\)
\(C=\frac{99}{100}\)
\(B=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{98.99}-\frac{1}{99.100}\\
=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\\
=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\\
=-\left(1-\frac{1}{100}\right)=\frac{-99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
1)xE{2;0} 2)abcd=a000+b00+c0+d=a.1000+b.100+c.10+d=(a.1000+b.96+c.8)+(4.b+2.c+d)=8.(a.125+b.12+c)+(d+2.c+4.b). vì 8 chia hết cho 8 =>8.(a.125+b.12+c) chia hết cho 8. Mà d+2.c+4.b chia hết cho 8. =>8.(a.125+b.12+c)+(d+2.c+4.b) chia hết cho 8 hay abcd chia hết cho 8. 3)3.S=1.2.3+2.3.3+3.4.3+...+99.100.3. =>3S=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98) =>3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100. =>3S=99.100.101.=>3s=979902=>S=326634.
Công thức :\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
Áp dụng:
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Vậy.................
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(D=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(D=1-\frac{1}{100}\)
\(D=\frac{99}{100}\)
kakaka thanks nhé mãi mới có người có ý thức
\(D=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(D=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
\(D=1-\frac{1}{100}=\frac{99}{100}\)
~ Hok tốt ~