$a$)

Ta có: $\dfrac{1}{33} > \dfrac{1}{36}$

           $\dfrac{1}{34} > \dfrac{1}{36}$

           $\dfrac{1}{35} > \dfrac{1}{36}$

$⇒ M = \dfrac{1}{33} + \dfrac{1}{34} + \dfrac{1}{35} + \dfrac{1}{36} > \dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} = \dfrac{1}{36} \times 4 = \dfrac{1}{9}$

Mà $\dfrac{1}{9} > \dfrac{1}{10}$

$⇒ M > \dfrac{1}{10}$

$b$) `A=1+1/2+1/3+1/4+1/5+....+1/{15} + 1/{16}`

`A = 1 + (1/2 + 1/3  + 1/4) + (1/5+1/6+1/7+1/8) + (1/9+1/{10}+1/{11}+1/{12})+ (1/{13} + 1/{14} + 1/{15} + 1/{16})`

`A > 1 + (1/2 + 1/3 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/{12} + 1/{12} + 1/{12} + 1/{12}) + (1/{16} + 1/{16} + 1/{16} + 1/{16})`

` A > 1 + (1/2 + 1/3 + 1/4) + 1/8 \times 4 + 1/{12} \times 4 + \frac{1}{16} \times 4`

`A > 1 + 2 \times (1/2+1/3+1/4)`

`A > 1 + 2 \times \frac{13}{12}`

`A > 1 + \frac{13}{6}`

`A > \frac{19}{6} > \frac{18}{6} =3`

$\text{1) M=$\dfrac{1}{33}$+....+$\dfrac{1}{36}$}$

$\text{Xét M:$\dfrac{1}{33}$> $\dfrac{1}{36}$}$

          $\text{.....(tất cả đều lớn hơn $\dfrac{1}{36}$)}$

          $\text{$\dfrac{1}{36}$= $\dfrac{1}{36}$}$

$\text{⇒M>4×$\dfrac{1}{36}$=$\dfrac{1}{9}$}$ 

$⇒M>N$

$\text{2) A=1+$\dfrac{1}{2}$+...+ $\dfrac{1}{16}$}$

$\text{⇒A=1+$\dfrac{1}{2}$+( $\dfrac{1}{3}$+$\dfrac{1}{4}$)+( $\dfrac{1}{5}$+..+ $\dfrac{1}{8}$)+($\dfrac{1}{9}$+..+$\dfrac{1}{12}$)+(  $\dfrac{1}{13}$+..+ $\dfrac{1}{16}$)}$

$\text{⇒A=1+ $\dfrac{1}{2}$+S1+S2+S3+S4}$

$\text{Xét S1: $\dfrac{1}{3}$>$\frac{1}{4}$}$  

            $\text{$\dfrac{1}{4}$=$\frac{1}{4}$}$  

$\text{⇒S1>2×$\dfrac{1}{4}$=$\dfrac{1}{2}$}$ 

$\text{Xét S2:$\dfrac{1}{5}$> $\dfrac{1}{8}$}$

          $\text{......(tất cả đều lớn hơn $\dfrac{1}{8}$}$ 

          $\text{$\dfrac{1}{8}$= $\dfrac{1}{8}$}$ 

$\text{⇒S2>4×$\dfrac{1}{8}$=$\dfrac{1}{2}$}$

$\text{Xét S3:$\dfrac{1}{9}$>$\dfrac{1}{12}$}$  

          $\text{......(tất cả đều lớn hơn $\dfrac{1}{12}$}$ 

            $\text{$\dfrac{1}{12}$=$\dfrac{1}{12}$}$

$\text{⇒S3> $\dfrac{1}{12}$×4= $\dfrac{1}{3}$}$ 

$\text{Xét S4:$\dfrac{1}{13}$> $\dfrac{1}{16}$}$ 

            $\text{......(tất cả đều lớn hơn $\dfrac{1}{16}$)}$ 

          $\text{$\dfrac{1}{16}$=$\dfrac{1}{16}$}$

$\text{⇒S4>  $\dfrac{1}{16}$ ×4=$\dfrac{1}{4}$}$ 

$\text{Do đó A=1+$\dfrac{1}{2}$+S1+S2+S3+S4>1+$\dfrac{1}{2}$+$\dfrac{1}{2}$+ $\dfrac{1}{2}$+$\dfrac{1}{3}$+$\dfrac{1}{4}$}$

$\text{⇒A> $\dfrac{37}{12}$> $\dfrac{36}{12}$=3}$

$⇒A>B$