Xét : $\left(a^2+b^2+c^2+d^2\right)-\left(a+b+c+d\right)$

        $=a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)+d\left(d-1\right)$

Vì $a$ là  số nguyên dương nên $a,\left(a-1\right)$ là hai số tự nhiên liên tiếp . 

$\Rightarrow a\left(a-1\right)$ chia hết cho 2. Tương tự ta có : $b\left(b-1\right);c\left(c-1\right);d\left(d-1\right)$ đều chia hết cho 2.

$\Rightarrow a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)+d\left(d-1\right)$ là số chẵn . 

Lại có : $a^2+c^2=b^2+d^2\Rightarrow a^2+b^2+c^2+d^2=2\left(b^2+d^2\right)$ là số chẵn .

Do đó : $a+b+c+d$ là số chẵn mà $a+b+c+d>2$ (Do $a,b,c,d\in N^{\ast }$)

Vậy : $a+b+c+d$ là hợp số .

x=11+11-11+22-30+7

b: $A=\genfrac{}{}{0ex}{}{2}{3}\left(\genfrac{}{}{0ex}{}{1}{60}-\genfrac{}{}{0ex}{}{1}{63}+\genfrac{}{}{0ex}{}{1}{63}-\genfrac{}{}{0ex}{}{1}{66}+...+\genfrac{}{}{0ex}{}{1}{117}-\genfrac{}{}{0ex}{}{1}{120}\right)+\genfrac{}{}{0ex}{}{2}{2003}$

$=\genfrac{}{}{0ex}{}{2}{3}\cdot \genfrac{}{}{0ex}{}{1}{120}+\genfrac{}{}{0ex}{}{2}{2003}$

$=2\left(\genfrac{}{}{0ex}{}{1}{360}+\genfrac{}{}{0ex}{}{1}{2003}\right)$

$B=\genfrac{}{}{0ex}{}{5}{4}\left(\genfrac{}{}{0ex}{}{1}{40}-\genfrac{}{}{0ex}{}{1}{44}+\genfrac{}{}{0ex}{}{1}{44}-\genfrac{}{}{0ex}{}{1}{48}+...+\genfrac{}{}{0ex}{}{1}{76}-\genfrac{}{}{0ex}{}{1}{80}\right)+\genfrac{}{}{0ex}{}{5}{2003}$

$=\genfrac{}{}{0ex}{}{5}{4}\cdot \genfrac{}{}{0ex}{}{1}{80}+\genfrac{}{}{0ex}{}{5}{2003}$

$=5\left(\genfrac{}{}{0ex}{}{1}{320}+\genfrac{}{}{0ex}{}{1}{2003}\right)$

Vì 1/360+1/2003<1/320+1/2003

nên A<B 

b: $A=\genfrac{}{}{0ex}{}{2}{3}\left(\genfrac{}{}{0ex}{}{1}{60}-\genfrac{}{}{0ex}{}{1}{63}+\genfrac{}{}{0ex}{}{1}{63}-\genfrac{}{}{0ex}{}{1}{66}+...+\genfrac{}{}{0ex}{}{1}{117}-\genfrac{}{}{0ex}{}{1}{120}\right)+\genfrac{}{}{0ex}{}{2}{2003}$

$=\genfrac{}{}{0ex}{}{2}{3}\cdot \genfrac{}{}{0ex}{}{1}{120}+\genfrac{}{}{0ex}{}{2}{2003}$

$=2\left(\genfrac{}{}{0ex}{}{1}{360}+\genfrac{}{}{0ex}{}{1}{2003}\right)$

$B=\genfrac{}{}{0ex}{}{5}{4}\left(\genfrac{}{}{0ex}{}{1}{40}-\genfrac{}{}{0ex}{}{1}{44}+\genfrac{}{}{0ex}{}{1}{44}-\genfrac{}{}{0ex}{}{1}{48}+...+\genfrac{}{}{0ex}{}{1}{76}-\genfrac{}{}{0ex}{}{1}{80}\right)+\genfrac{}{}{0ex}{}{5}{2003}$

$=\genfrac{}{}{0ex}{}{5}{4}\cdot \genfrac{}{}{0ex}{}{1}{80}+\genfrac{}{}{0ex}{}{5}{2003}$

$=5\left(\genfrac{}{}{0ex}{}{1}{320}+\genfrac{}{}{0ex}{}{1}{2003}\right)$

Vì 1/360+1/2003<1/320+1/2003

nên A<B

C1:Ta co

+)A=2/60*63+2/63*66+...+2/117*120+2/2003

A*3/2=3/60*63+3/63*66+...+3/117*120+3/2003

A*3/2=1/60-1/63+1/63-1/66+...+1/117-1/120+3/2003

A*3/2=1/60-1/120+3/2003

A=(1/120+3/2003)*2/3

+)B=5/40*44+5/44*48+...+5/76*80+5/2003

B*4/5=4/40*44+4/44*48+...+4/76*80+4/2003

B*4/5=1/40-1/44+1/44-1/48+...+1/76-1/80+4/2003

B*4/5=1/40-1/80+4/2003

B=(1/80+4/2003)*5/4

Tu tren ta co A=(1/120+3/2003)*2/3

B=(1/80+4/2003)*5/4

Vay A<B(Vi 1/120<1/80;3/2003<4/2003;2/3<5/4)

C2:

b: $A=$$\genfrac{}{}{0ex}{}{2}{3}\left(\genfrac{}{}{0ex}{}{1}{60}-\genfrac{}{}{0ex}{}{1}{63}+\genfrac{}{}{0ex}{}{1}{63}-\genfrac{}{}{0ex}{}{1}{66}+...+\genfrac{}{}{0ex}{}{1}{117}-\genfrac{}{}{0ex}{}{1}{120}\right)+\genfrac{}{}{0ex}{}{2}{2003}$

$=\genfrac{}{}{0ex}{}{2}{3}\cdot \genfrac{}{}{0ex}{}{1}{120}+\genfrac{}{}{0ex}{}{2}{2003}$

$=2\left(\genfrac{}{}{0ex}{}{1}{360}+\genfrac{}{}{0ex}{}{1}{2003}\right)$

$B=\genfrac{}{}{0ex}{}{5}{4}\left(\genfrac{}{}{0ex}{}{1}{40}-\genfrac{}{}{0ex}{}{1}{44}+\genfrac{}{}{0ex}{}{1}{44}-\genfrac{}{}{0ex}{}{1}{48}+...+\genfrac{}{}{0ex}{}{1}{76}-\genfrac{}{}{0ex}{}{1}{80}\right)+\genfrac{}{}{0ex}{}{5}{2003}$

$=\genfrac{}{}{0ex}{}{5}{4}\cdot \genfrac{}{}{0ex}{}{1}{80}+\genfrac{}{}{0ex}{}{5}{2003}$

$=5\left(\genfrac{}{}{0ex}{}{1}{320}+\genfrac{}{}{0ex}{}{1}{2003}\right)$

Vì 1/360+1/2003<1/320+1/2003

nên A<B

+)A=2/60*63+2/63*66+...+2/117*120+2/2003

A*3/2=3/60*63+3/63*66+...+3/117*120+3/2003

A*3/2=1/60-1/63+1/63-1/66+...+1/117-1/120+3/2003

A*3/2=1/60-1/120+3/2003

A=(1/120+3/2003)*2/3

+)B=5/40*44+5/44*48+...+5/76*80+5/2003

B*4/5=4/40*44+4/44*48+...+4/76*80+4/2003

B*4/5=1/40-1/44+1/44-1/48+...+1/76-1/80+4/2003

B*4/5=1/40-1/80+4/2003

B=(1/80+4/2003)*5/4

Tu tren ta co A=(1/120+3/2003)*2/3

B=(1/80+4/2003)*5/4

Vay A<B(Vi 1/120<1/80;3/2003<4/2003;2/3<5/4)