Lời giải:

Ta có:

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

Vậy ta có đpcm.

A = \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)

Vậy: A = \(\dfrac{49}{100}\)

A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

A=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A=\(\dfrac{1}{2}-\dfrac{1}{100}\)

A=\(\dfrac{49}{100}\)

\(\dfrac{49}{100}\)

d, `3,15+2,4=5,55`

e, \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\dfrac{11}{11}=\dfrac{5}{7}.1=\dfrac{5}{7}\)

f, `1,25.3,6+3,6.8,75=3,6(1,25+8,75)=3,6.10=36`

\(h,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

\(e\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}\times1=\dfrac{5}{7}\)

\(f3.6\times\left(1.25+8.75\right)=3.6\times10=36\)

\(a,\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}.\)

\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}.\)

\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{1}{2}.\dfrac{12}{5}.\)

\(=-\dfrac{3}{4}+\dfrac{12}{10}.\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}.\)

\(=-\dfrac{15}{20}+\dfrac{24}{20}=\dfrac{9}{20}.\)

Vậy.....

\(b,\left(-2\right)^2-1\dfrac{5}{27}.\left(-\dfrac{3}{2}\right)^3.\)

\(=4-1\dfrac{5}{27}.\left(-\dfrac{27}{8}\right).\)

\(=4-\dfrac{32}{27}.\left(-\dfrac{27}{8}\right).\)

\(=4-\left(-4\right).\)

\(=4+4=8.\)

Vậy.....

\(c,\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}.\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}+0+0+...+0-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}-\dfrac{1}{100}.\)

\(=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}.\)

Vậy.....

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\\ =1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+......+\left(-\dfrac{1}{99}+\dfrac{1}{99}\right)-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{100-1}{100}=\dfrac{99}{100}\)

Ta có : 

\(\dfrac{1}{1.2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)

\(...\)

\(\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow\) biểu thức chỉ còn :

\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

=1/1-1/100

=100/100-1/100

=99/100