a) O nằm giữa A và B nên:

\(AB=OA+OB\)

\(\Rightarrow OA=1,5+3=4,5\left(cm\right)\)

b) C nằm giữa O và B 

\(OB=OC+BC\)

\(\Rightarrow BC=OB-OC\)

\(\Rightarrow BC=3-1,5=1,5\left(cm\right)\)

\(OC=BC=1,5\left(cm\right)\)

TH1: với \(a>b\) 

\(\dfrac{a}{b}-1=\dfrac{a-b}{b}\)

\(\dfrac{a+n}{b+n}-1=\dfrac{a+n-\left(b+n\right)}{b+n}=\dfrac{a-b}{b+n}\)

Mà: \(b+n>b\)

\(\Rightarrow\dfrac{a-b}{b+n}< \dfrac{a-b}{b}\)

\(\Rightarrow\dfrac{a+n}{b+n}-1< \dfrac{a}{b}-1\)

\(\Rightarrow\dfrac{a+n}{b+n}< \dfrac{a}{b}\)

TH2: với \(a< b\)

\(1-\dfrac{a}{b}=\dfrac{b-a}{b}\)

\(1-\dfrac{a+n}{b+n}=\dfrac{\left(b+n\right)-\left(a+n\right)}{b+n}=\dfrac{b-a}{b+n}\)

Mà: \(b+n>b\)

\(\Rightarrow\dfrac{b-a}{b+n}< \dfrac{b-a}{b}\)

\(\Rightarrow1-\dfrac{a+n}{b+n}< 1-\dfrac{a}{b}\)

\(\Rightarrow\dfrac{a+n}{b+n}>\dfrac{a}{b}\) 

TH3: \(a=b\)

\(\dfrac{a+n}{b+n}=\dfrac{a+n}{a+n}=1\)

\(\dfrac{a}{b}=\dfrac{a}{a}=1\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{a+n}{b+n}=1\)

\(S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}\\ 2S=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\\ 2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}\right)\\ S=1-\dfrac{1}{512}=\dfrac{511}{512}\)

Lời giải:

$S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}$

$2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^8}$

$\Rightarrow 2S-S=1-\frac{1}{2^9}$

$\Rightarrow S=1-\frac{1}{2^9}$

\(S=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\left(1-\dfrac{1}{5^2}\right)\left(1-\dfrac{1}{6^2}\right)...\left(1-\dfrac{1}{99^2}\right)\\ =\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{9801}\right)\\ =\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{9800}{9801}\\ =\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{98.100}{99.99}\\ =\dfrac{1.2.3...98}{2.3.4...99}.\dfrac{3.4.5...100}{2.3.4...99}\\ =\dfrac{1}{99}.\dfrac{100}{2}=\dfrac{50}{99}\)

Lời giải:

\(S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{98.100}{99^2}\\ =\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{98.100}{99^2}\\ =\frac{1.3.2.4.3.5....98.100}{2^2.3^2.4^2...99^2}\\ =\frac{(1.2.3...98)(3.4.5..100)}{(2.3.4...99)(2.3.4...99)}\)

\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5..100}{2.3.4...99}=\frac{1}{99}.\frac{100}{2}=\frac{100}{198}\)

\(S=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=\dfrac{101}{2}\)

\(S=1+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+...+\dfrac{649}{650}\)

\(=1+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+...+1-\dfrac{1}{650}\)

\(=\left(1+1+...+1\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{650}\right)\)

\(=\left(1+1+...+1\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{25\cdot26}\right)\)

\(=25-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{25}-\dfrac{1}{26}\right)\)

\(=25-\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=25-\dfrac{12}{26}=25-\dfrac{6}{13}=\dfrac{319}{13}\)

Sửa đề: \(1+\dfrac{1}{8}+\dfrac{1}{24}+...+\dfrac{4}{x\left(x+4\right)}=\dfrac{99}{100}\)

=>\(\dfrac{4}{4}+\dfrac{4}{32}+\dfrac{4}{96}+...+\dfrac{4}{x\left(x+4\right)}==\dfrac{99}{100}\)

=>\(\dfrac{4}{4}+\dfrac{4}{4\cdot8}+\dfrac{4}{8\cdot12}+...+\dfrac{4}{x\left(x+4\right)}=\dfrac{99}{100}\)
=>\(\dfrac{4}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{99}{100}\)

=>\(\dfrac{5}{4}-\dfrac{1}{x+4}=\dfrac{99}{100}\)

=>\(\dfrac{1}{x+4}=\dfrac{125-99}{100}=\dfrac{26}{100}=\dfrac{13}{50}\)

=>\(x+4=\dfrac{50}{13}\)

=>\(x=-\dfrac{2}{13}\)

Lời giải:
$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}$

$=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$

$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50})$

$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$

$\Rightarrow (\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}):(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100})=1$