\(\overline{abcd}+\overline{dcba}=11220\)

\(\Leftrightarrow1000a+100b+10c+d+1000d+100c+10d+a=11220\)

\(\Leftrightarrow1001\left(a+d\right)+110\left(b+c\right)=11200\)

\(\Leftrightarrow11\cdot91\left(a+d\right)=110\left(102-\left(b+c\right)\right)\)

\(\Leftrightarrow91\left(a+d\right)=10\left(102-\left(b+c\right)\right)\)(1)

=> 102 - (b + c) chia hết cho 91 mà 0 <= b + c <= 18 => 84 <= 102 - (b + c) <=102. Trong khoảng [84;102] chỉ có 91 là bội của 91.

Do đó: 102 - (b + c) = 91 => b + c = 11

Thay vào (1) => a + b = 10

Vậy, tổng a + b + c + d = 10 + 11 = 21.

\(\left(3x+2\right)^2-\left(x+2\right)\left(9x-1\right)\)

\(=\left(9x^2+6x+4\right)-\left(9x^2-x+18x-2\right)\)

\(=9x^2+6x+4-9x^2-17x+2\)

\(=6-11x\)

(a+b-c)-(b-c+d)

= a+b-c-b+c-d

= a+(b-b)+(c-c)+d

= a+0+0+d

=a+d

\(\left(117+30\right)\cdot x=882:2\)

\(147\cdot x=441\)

\(x=441:147\)

\(x=3.\)