b: BCNN(21;55)=1155

a)        \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)

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

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

b)       \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)\(=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

     \(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{110}\)

       \(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

c)   \(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{97.99}\)   \(=\frac{13-11}{11.13}+\frac{15-13}{13.15}+\frac{17-15}{15.17}+...+\frac{99-97}{97.99}\)  

\(=\frac{1}{11}+\frac{1}{13}-\frac{1}{13}+\frac{1}{15}-\frac{1}{15}+\frac{1}{17}...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{11}-\frac{1}{99}=\frac{8}{99}\)

k biet l

1/1 * 2 + 1/2 *3  + 1/3 * 4 +....... + 1/99 * 100

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

= 1 - 1/100 = 99/100

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Ôi cảm ơn ban nhiều lắm

\(P=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{x-9}\)

\(=\dfrac{x-\sqrt{x}-6}{x-9}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{x-9}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

4:

a: P>4/5

=>P-4/5>0

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{4}{5}>0\)

=>\(\dfrac{5\sqrt{x}+10-4\sqrt{x}-12}{5\sqrt{x}+15}>0\)

=>\(\sqrt{x}-2>0\)

=>x>4

b: \(P>\dfrac{2\sqrt{x}}{5}\)

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2\sqrt{x}}{5}>0\)

=>\(\dfrac{5\sqrt{x}+10-2x-6\sqrt{x}}{5\sqrt{x}+15}>0\)

=>\(-2x-\sqrt{x}+10>0\)

=>\(-2x-5\sqrt{x}+4\sqrt{x}+10>0\)

=>\(\left(2\sqrt{x}+5\right)\left(-\sqrt{x}+2\right)>0\)

=>\(-\sqrt{x}+2>0\)

=>0<=x<4

5:

a: \(P-\dfrac{1}{2}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{1}{2}\)

\(=\dfrac{2\sqrt{x}+4-\sqrt{x}-3}{2\sqrt{x}+6}=\dfrac{\sqrt{x}+1}{2\sqrt{x}+6}>0\)

=>P>1/2

b: \(P-1=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-1=\dfrac{\sqrt{x}+2-\sqrt{x}-3}{\sqrt{x}+3}\)

\(=\dfrac{-1}{\sqrt{x}+3}< 0\)

\(P^2-P=P\left(P-1\right)\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\cdot\dfrac{-1}{\sqrt{x}+3}< 0\)

=>P^2<P

=>P>P^2