Упр.1100 ГДЗ Мерзляк Полонский 6 класс (Математика)
Раскройте скобки:
- $$-12\left(\frac{5}{6}a-\frac{1}{4}b+\frac{7}{24}c-\frac{1}{12}\right)$$;
- $$\left(16a+8b-\frac{5}{9}x-\frac{4}{9}d\right)\left(-\frac{9}{32}n\right)$$;
- $$-\frac{4}{15}bc\left(-45a-30d+3\cdot\frac{3}{4}m-\frac{3}{8}\right)$$;
- $$\left(-3{,}6ab+20a-b-100\right)(-5xy)$$.
Раскроем скобки, умножая каждый член в скобках на вынесенный множитель.
1) $$-12\left(\frac{5}{6}a-\frac{1}{4}b+\frac{7}{24}c-\frac{1}{12}\right)$$
$$-12\cdot \frac{5}{6}a=-10a,\quad -12\cdot \left(-\frac{1}{4}b\right)=3b,\quad -12\cdot \frac{7}{24}c=-\frac{7}{2}c,\quad -12\cdot \left(-\frac{1}{12}\right)=1$$
$$-12\left(\frac{5}{6}a-\frac{1}{4}b+\frac{7}{24}c-\frac{1}{12}\right)=-10a+3b-\frac{7}{2}c+1$$
2) $$\left(16a+8b-\frac{5}{9}x-\frac{4}{9}d\right)\cdot \left(-\frac{9}{32}n\right)$$
$$16a\cdot \left(-\frac{9}{32}n\right)=-\frac{9}{2}an,\quad 8b\cdot \left(-\frac{9}{32}n\right)=-\frac{9}{4}bn,$$
$$-\frac{5}{9}x\cdot \left(-\frac{9}{32}n\right)=\frac{5}{32}xn,\quad -\frac{4}{9}d\cdot \left(-\frac{9}{32}n\right)=\frac{1}{8}dn$$
$$\left(16a+8b-\frac{5}{9}x-\frac{4}{9}d\right)\cdot \left(-\frac{9}{32}n\right) =-\frac{9}{2}an-\frac{9}{4}bn+\frac{5}{32}xn+\frac{1}{8}dn$$
3) $$-\frac{4}{15}bc\left(-45a-30d+3\frac{3}{4}m-\frac{3}{8}\right)$$
Сначала представим смешанное число: $$3\frac{3}{4}=\frac{15}{4}$$.
$$-\frac{4}{15}bc\cdot (-45a)=12abc,\quad -\frac{4}{15}bc\cdot (-30d)=8bcd,$$
$$-\frac{4}{15}bc\cdot \frac{15}{4}m=-bcm,\quad -\frac{4}{15}bc\cdot \left(-\frac{3}{8}\right)=\frac{1}{10}bc$$
$$-\frac{4}{15}bc\left(-45a-30d+3\frac{3}{4}m-\frac{3}{8}\right) =12abc+8bcd-bcm+\frac{1}{10}bc$$
4) $$(-3{,}6ab+20a-b-100)\cdot (-5xy)$$
$$-3{,}6ab\cdot (-5xy)=18abxy,\quad 20a\cdot (-5xy)=-100axy,$$
$$(-b)\cdot (-5xy)=5bxy,\quad (-100)\cdot (-5xy)=500xy$$
$$(-3{,}6ab+20a-b-100)\cdot (-5xy)=18abxy-100axy+5bxy+500xy$$
Ответ
1) $$-10a+3b-\frac{7}{2}c+1$$
2) $$-\frac{9}{2}an-\frac{9}{4}bn+\frac{5}{32}xn+\frac{1}{8}dn$$
3) $$12abc+8bcd-bcm+\frac{1}{10}bc$$
4) $$18abxy-100axy+5bxy+500xy$$
