Упр.1100 ГДЗ Мерзляк Полонский 6 класс (Математика)

1) -12 * (5/6*a — 1/4*b + 7/24*c — 1/12);
2) (16a + 8b — 5/9*x — 4/9*d) * (-9/32*n);
3) -4/15*bc * (-45a — 30d + 3*3/4*m — 3/8);
4) (-3,6ab + 20a — b — 100) * (-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) $$18abxy-100axy+5bxy+500xy$$