Упр.271 ГДЗ Никольский Потапов 7 класс (Алгебра)
- 7. Упростите:
а) $$\left(2a^2b-10b^3\right)-\left(4a^2b-12b^3\right);$$
б) $$\left(3xy^2+7x^2y\right)-\left(2xy^2-6x^2y\right);$$
в) $$12ab-30bc-3cx-\left(15bc+9cx\right);$$
г) $$\left(10abc-8bcx-21cxy\right)-\left(-6abc+bcx-cxy\right);$$
д) $$\left(0{,}6ab-0{,}5bc+cx\right)-\left(2{,}5bc-0{,}5ab-cx\right);$$
е) $$\left(\frac{1}{2}x^2y^2-\frac{2}{3}x^2y^2-\frac{5}{6}a^2b\right)-\left(a^2b-\frac{1}{3}x^2y^2+\frac{1}{12}ab-\frac{1}{4}\right).$$
Раскроем скобки, учитывая знаки перед ними, и приведём подобные слагаемые.
$$(2a^2b-10b^3)-(4a^2b-12b^3)=2a^2b-10b^3-4a^2b+12b^3$$
$$=-2a^2b+2b^3=2b^3-2a^2b$$
$$(3xy^2+7x^2y)-(2xy^2-6x^2y)=3xy^2+7x^2y-2xy^2+6x^2y$$
$$=xy^2+13x^2y$$
$$12ab-30bc-3cx-(15bc+9cx)=12ab-30bc-3cx-15bc-9cx$$
$$=12ab-45bc-12cx$$
$$(10abc-8bcx-21cxy)-(-6abc+bcx-cxy)$$
$$=10abc-8bcx-21cxy+6abc-bcx+cxy$$
$$=16abc-9bcx-20cxy$$
$$(0{,}6ab-0{,}5bc+cx)-(2{,}5bc-0{,}5ab-cx)$$
$$=0{,}6ab-0{,}5bc+cx-2{,}5bc+0{,}5ab+cx$$
$$=1{,}1ab-3bc+2cx$$
$$\left(\frac12x^2y^2-\frac23x^2y^2-\frac56a^2b\right)-\left(a^2b-\frac13x^2y^2+\frac1{12}ab-\frac14\right)$$
$$=\frac12x^2y^2-\frac23x^2y^2-\frac56a^2b-a^2b+\frac13x^2y^2-\frac1{12}ab+\frac14$$
$$=\left(\frac12-\frac23+\frac13\right)x^2y^2-\left(\frac56+1\right)a^2b-\frac1{12}ab+\frac14$$
$$=\frac16x^2y^2-\frac{11}{6}a^2b-\frac1{12}ab+\frac14$$
Ответ
а) $$2b^3-2a^2b$$; б) $$xy^2+13x^2y$$; в) $$12ab-45bc-12cx$$; г) $$16abc-9bcx-20cxy$$; д) $$1{,}1ab-3bc+2cx$$; е) $$\frac16x^2y^2-\frac{11}{6}a^2b-\frac1{12}ab+\frac14$$.
