Ta có:

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

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

\(\dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}\)

...

\(\dfrac{1}{9^2}=\dfrac{1}{9\cdot9}< \dfrac{1}{8\cdot9}\)

\(\dfrac{1}{10^2}=\dfrac{1}{10\cdot10}< \dfrac{1}{9\cdot10}\)

\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A< 1-\dfrac{1}{10}\)

\(\Rightarrow A< \dfrac{9}{10}\)

\(\Rightarrow A< 1\) (vì: \(\dfrac{9}{10}< 1\))

221​=2⋅21​<1⋅21​

132=13⋅3<12⋅3321​=3⋅31​<2⋅31​

142=14⋅4<13⋅4421​=4⋅41​<3⋅41​

...

192=19⋅9<18⋅9921​=9⋅91​<8⋅91​

1102=110⋅10<19⋅101021​=10⋅101​<9⋅101​

⇒�=122+132+142+...+1102<11⋅2+12⋅3+13⋅4+...+19⋅10⇒A=221​+321​+421​+...+1021​<1⋅21​+2⋅3

a) Đ

b) S 

\(\dfrac{7}{10}-\dfrac{1}{5} \\ =\dfrac{7}{10}-\dfrac{2}{10}\\ =\dfrac{5}{10}=\dfrac{1}{2}\)

c) S

\(\dfrac{5}{4}+\dfrac{5}{12}\\ =\dfrac{15}{12}+\dfrac{5}{12}\\ =\dfrac{20}{12}=\dfrac{5}{3}\)

d) Đ

1: 

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

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

...

\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)

=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)

=>\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\)

=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)

=>\(A>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

=>\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

=>\(A>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{5}{10}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

Do đó: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)

\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\)  - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)

\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)

\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\) 

\(\dfrac{3}{16}\)  - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)

 ( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\) 

    \(\dfrac{41}{16}\)              - \(x\)    = \(\dfrac{5}{2}\)

                         \(x\)    = \(\dfrac{41}{16}\)  - \(\dfrac{5}{2}\)

                         \(x\)     = \(\dfrac{1}{16}\)

2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\)) 

    \(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)

   - \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)

    - \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)

       \(x\)   = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)

       \(x\)    = \(\dfrac{5}{6}\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`

`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`

`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`

`=> 3/16 - x + 19/8 = 2 1/2`

`=> 3/16 - x = 2 1/2 - 19/8`

`=> 3/16 - x =1/8`

`=> x = 3/16 - 1/8`

`=> x = 1/16`

Vậy, `x = 1/16`

`2,`

`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`

`=> 1/2 * (-11/15) = 1/5 - x + 4/15`

`=> -11/30 = x + 1/5 - 4/15`

`=> x + (-1/15) = -11/30`

`=> x = -11/30 + 1/15`

`=> x = -3/10`

Vậy, `x = -3/10.`

Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)

\(\Rightarrow2^2A=1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{98}}\)

\(\Rightarrow2^2A-A=\left(1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\right)\)

\(\Rightarrow3A=1-\dfrac{1}{2^{100}}\)

\(\Rightarrow A=\dfrac{1-\dfrac{1}{2^{100}}}{3}< \dfrac{1}{3}\)(đpcm)