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Số số hạng là(99,100-10,11):0,01+1=8900
=>Tổng=8900:2x(10,11+99,100)=485984,5
A= 5/10.11+5/11.12+5/12.13+5/13.14+5/14.15
A= (1/10.11+1/11.12+1/12.13+1/13.14+1/14.15) :5
A= [(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+(1/13-1/14)+(1/14-1/15)] :5
A= (1/10-1/15):5
A= 1/30:5
A= 1/150
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(\Rightarrow A=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(\Rightarrow A=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(\Rightarrow A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(\Rightarrow A=7.\frac{3}{35}\)
\(\Rightarrow A=\frac{3}{5}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}\)
\(\Leftrightarrow C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(\Leftrightarrow C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(\Leftrightarrow C=7\cdot\frac{3}{35}\)
\(\Leftrightarrow C=\frac{3}{5}\)
C=7(1/10.11+1/11.12+...+1/69.70)
=7(1/10-1/11+1/11-1/12+...+1/69-1/70)
=7(1/10-1/70)
=7.3/35
=3/5
Đặt A=1.2+2.3+...+99.100
A.3=1.2.3+2.3.3+...+99.100.3
A.3=1.2.[3-0]+2.3.[4-1]+...+99.100.[101-98]
A.3=1.2.3+2.3.4-1.2.3+...+99.100.101-99.100.98
A.3=99.100.101
A.3=999900
A=333300
A = 1.2+2.3+3.4+4.5+...+98.99+99.100
3A = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3A = 99.100.101
3A = 999900
A = 333300
nhấn đúng cho mk nha!!!!!!!!!!!!
a)=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
b)\(=\frac{201.204+1}{\left(201+2\right).204-407}\)
\(=\frac{201.204+1}{201.204+2.204-407}\)
\(=\frac{201.204+1}{201.204+1}\)
=1
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Đặt : \(A=10.11+11.12+...+98.99+99.100\)
\(\Rightarrow3A=10.11.3+11.12.3+...+98.99.3+99.100.3\)
\(\Rightarrow3A=10.11.\left(12-9\right)+11.12.\left(13-10\right)+...+\)\(98.99.\left(100-97\right)+99.100.\left(101-98\right)\)