Задание 271

Решение:

• $${\largeа)}\ (2a^2b-10b^3)-(4a^2b-12b^3)=2a^2b-10b^3-4a^2b+12b^3=(2-4)a^2b+(-10+12)b^3=(-2)a^2b+2b^3$$

• $${\largeб)}\ (3xy^2+7x^2y)-(2xy^2-6x^2y)=3xy^2+7x^2y-2xy^2+6x^2y=(3-2)xy^2+(7+6)x^2y=xy^2+13x^2y$$

• $${\largeв)}\ 12ab-30bc-3cx-(15bc+9cx)=12ab-30bc-3cx-15bc-9cx=12ab+(-30-15)bc+(-3-9)cx=12ab-45bc-12cx$$

• $${\largeг)}\ (10abc-8bcx-21cxy)-(-6abc+bcx-cxy)=10abc-8bcx-21cxy+6abc-bcx+cxy=(10+6)abc+(-8-1)bcx+(-21+1)cxy=16abc-9bcx-20cxy$$

• $${\largeд)}\ (0{,}6ab-0{,}5bc+cx)-(2{,}5bc-0{,}5ab-cx)=0{,}6ab-0{,}5bc+cx-2{,}5bc+0{,}5ab+cx=(0{,}6+0{,}5)ab+(-0{,}5-2{,}5)bc+(1+1)cx=1{,}1ab-3bc+2cx$$

• $${\largeе)}\ \left(\frac{1}{2}x^2y^2-\frac{2}{3}ab-\frac{5}{6}a^2b-1\right)-\left(a^2b-\frac{1}{3}x^2y^2+\frac{1}{12}ab-\frac{1}{4}\right)=\frac{1}{2}x^2y^2-\frac{2}{3}ab-\frac{5}{6}a^2b-1-a^2b+\frac{1}{3}x^2y^2-\frac{1}{12}ab+\frac{1}{4}=\left(\frac{3}{6}+\frac{2}{6}\right)x^2y^2+\left(-\frac{8}{12}-\frac{1}{12}\right)ab+\left(-\frac{5}{6}-1\right)a^2b+\left(-\frac{4}{4}+\frac{1}{4}\right)=\frac{5}{6}x^2y^2-\frac{9}{12}ab-1\frac{5}{6}a^2b-\frac{3}{4}=\frac{5}{6}x^2y^2-\frac{3}{4}ab-1\frac{5}{6}a^2b-\frac{3}{4}$$