Question

соч умаляю 3 и 5 прашу умаляю

Answer 1

Объяснение:

$3.\ a^2b-3ab+6a-18+ab-3b=(a^2b-3ab)+(6a-18)+(ab-3b)=\\=ab*(a-3)+6*(a-3)+b*(a-3)=(a-3)*(ab+6+b).\\5.\ (\frac{1}{6})^{-2}-(-\frac{2}{3})^0+(\frac{1}{4})^2:3=6^2-1+\frac{1}{16} :3 =36-1+\frac{1}{48} =35\frac{1}{48}.$

Answer 2

3.

а²b-3ab+6a-18+ab-3b

a²b-2ab+6a-18-3b

5.

(1/6)-²-(-2/3)0+(1/4)²:3

6²-1+1/16*1/3

36-1+1/48

35+1/48

1681/48