A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\)

A = \(\dfrac{1}{2^2}\)  + \(\dfrac{1}{2^2}\).(\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{50^2}\))

A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^2}\).(\(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + ... + \(\dfrac{1}{100.100}\))

A < \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\).(\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{49.50}\))

A < \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) .(\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)

A < \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) ( 1 - \(\dfrac{1}{50}\))

A < \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{50}\) 

A < \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\) < \(\dfrac{1}{2}\) (đpcm)

\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)

=>\(A=\dfrac{1}{2^2}+\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{2^2}+\dfrac{1}{2^2}\cdot\dfrac{49}{50}=\dfrac{1}{4}\left(1+\dfrac{49}{50}\right)=\dfrac{1}{4}\cdot\dfrac{99}{50}=\dfrac{99}{200}< \dfrac{1}{2}\)

   1 - 2  - 3  - 4 

= 1 - (2 + 3 + 4)

= 1  - (5 + 4)

= 1  - 9

= -(9 - 1)

= - 8 

\(Q=\left(n-2\right)\left(n-3\right)-\left(n-3\right)\left(n+2\right)\)

\(=\left(n-3\right)\left(n-2-n-2\right)\)

\(=-4\left(n-3\right)⋮2\)

=>Q là số chẵn

Q= (n-3)[n-2-(n+2)
   = (n-3) x (-4)
vì -4 chẵn nên Q chẵn ( đpcm )

Ta được 123000.

123000

nếu axd=cxb

\(\dfrac{a}{b}=\dfrac{c}{d}\) nếu ad=bc

\(\dfrac{-3}{5}+\dfrac{43}{10}+\dfrac{28}{24}-\dfrac{28}{15}\)

\(=\dfrac{-6+43}{10}+\dfrac{7}{6}-\dfrac{28}{15}\)

\(=\dfrac{37}{10}+\dfrac{7}{6}-\dfrac{28}{15}\)

\(=\dfrac{37\cdot3+7\cdot5-28\cdot2}{30}=\dfrac{90}{30}=3\)

\(\dfrac{-3}{5}+\dfrac{43}{10}+\dfrac{28}{24}-\dfrac{28}{15}\)

\(=\dfrac{37}{10}+\dfrac{28}{24}-\dfrac{28}{15}\)

\(=\dfrac{73}{15}-\dfrac{28}{15}\)

\(=3\)

=7/19.(8/11+3/11)

=7/19.1

=7/19

=7/19.(8/11+3/11)

=7/19.1

=7/19

=2/9-1/20-2/9

=(2/9-2/9)-1/20

=0-1/20

=-1/20

-1/20

-3/7 + 5/7 = 2/7

2/7