...

= 1/2-1/3+1/3-1/4+...+ 1/19-1/20

= 1/2-1/20

=9/20

có phải như thế này ko bn

\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}\)

A = \(\frac{9}{20}\)

\(B=\frac{1}{99.100}-\frac{1}{98.99}-\frac{1}{97.98}-.....-\frac{1}{1.2}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)

\(B=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)\)

B = \(-\frac{99}{100}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(A=\frac{1}{2}-\frac{1}{20}\)

\(A=\frac{10}{20}-\frac{1}{20}=\frac{9}{20}\)

Tương tự với ý B

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Đặt A = 1.2 + 2.3 + 3.4 + ... + 99.100

3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)

3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3A = 99.100.101

A = 33.100.101

A = 333300

\(A=1.2+2.3+3.4+4.5+...+97.98+98.99+99.100\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(A=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)

Ta có : A = 1/1.2 + 1/2.3 + .... + 1/98.99 + 1/99.100 .

=>       A = 1 - 1/2 + 1/2 - 1/3 + .... + 1/98 - 1/99 + 1/99 - 1/100 .

=>       A = 1 - 1/100 .

=>       A = 99/100 .

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)

Cảm ơn nhìu nha

10.11+11.12+12.13+...+97.98+98.99+99.100

=10-11+11-12+12-13+...+97-98+98-99+99-100

=10-100

=-90

Đặt A = 10.11 + 11.12 + ... + 98.99 + 99.100 

3A = 10.11.3 + 11.12.3 + ... + 98.99.3 + 99.100.3 

3A = 10.11.(12 -9) + 11.12.(13-10) + ... + 98.99.(100 - 97) + 99.100.(101-98)

3A = 10.11.12 - 9.10.11 + 11.12.13 - 10.11.12 + ... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100 

3A = (10.11.12 + 11.12.13 + ... + 98.99.100 + 99.100.101) - (9.10.11 + 10.11.12 + ... + 97.98.99 + 98.99.100)

3A = 99.100.101 - 9.10.11 

3A = 999799 

A = 999799 : 3  

ko spam

lèm rùi mè

=1/3 - 1/4 +1/4-1/5 +...+1/98 - 1/99

=1/3-1/99=32/99

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=1-\dfrac{1}{99}\)

\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)

Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)

Suy ra \(A>B\).

\(a,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{100}{100}-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(b,\frac{x}{y}=\frac{3}{5}\)

\(\Leftrightarrow\frac{x}{3}=\frac{y}{5}\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\)

\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{18}{8}=\frac{9}{4}\)

\(\Rightarrow\frac{x}{3}=\frac{9}{4}\Rightarrow x=\frac{27}{4}\)

\(\frac{y}{5}=\frac{9}{4}\Rightarrow y=\frac{45}{4}\)