№ 283 Алгебра = № 11.20 Математика
Розкрийте дужки та спростіть вираз:
Розв'язок:
1) х – (x – (2x – 3)) = х – (х – 2х + 3) = х – (–х + 3) = х + х – 3 = 2х – 3;
2) 5m – ((n – m) + 3n) = 5m – (n – m + 3n) = 5m – n + m – 3n = 6m – 4n;
3) 4p – (3p – (2p – (p + 1))) = 4p – (3p – (2p – p – 1)) = 4p – (3p – (p – 1)) = 4p – (3p – p + 1) = 4p – (2p + 1) = 4p – 2p – 1 = 2p – 1;
4) 5x – (2x – ((y – x) – 2y)) = 5x – (2x – (y – x – 2y)) = 5x – (2x – (–y – x)) = 5x – (2x + y + x) = 5x – (3x + y) = 5x – 3x – y = 2x – y;
5) $\frac{2}{3}\left(6a - \frac{3}{8}b\right) - \frac{2}{11}\left(4\frac{1}{8}a - 33b\right) =$
$= \frac{2}{3} \cdot 6a - \frac{2}{3} \cdot \frac{3}{8} \cdot b - \frac{2}{11} \cdot 4\frac{1}{8} \cdot a + \frac{2}{11} \cdot 33b =$
$= 4a - \frac{1}{4} \cdot b - \frac{3}{4} \cdot a + 6b =$
$= 3\frac{1}{4} \cdot a + 5\frac{3}{4} \cdot b$;
6) $-\frac{2}{9}(2{,}7m - 1{,}5n) + \frac{5}{6}(2n - 0{,}48m) =$
$= -\frac{2}{9} \cdot 2{,}7 \cdot m + \frac{2}{9} \cdot 1{,}5 \cdot n + \frac{5}{6} \cdot 2n - \frac{5}{6} \cdot 0{,}48m =$
$= -\frac{2}{9} \cdot \frac{27}{10} \cdot m + \frac{2}{9} \cdot \frac{3}{2} \cdot n + \frac{5}{6} \cdot 2n - \frac{5}{6} \cdot \frac{48}{100} \cdot m =$
$= -\frac{3}{5} \cdot m + \frac{1}{3} \cdot n + \frac{5}{3} \cdot n - \frac{2}{5} \cdot m =$
$= -m + 2n = 2n - m$.