Zapisz w jak...- Zadanie 6: Matematyka z plusem 8

Rozwiązanie

$a)$

$\frac{6 a + 18 b}{3} + \frac{- 12 a + 3}{9} = \frac{6 a + 18 b}{3} + \frac{3 \cdot ( - 4 a ) + 3 \cdot 1}{9} = \frac{6 a + 18 b}{3} + \frac{3 \cdot ( - 4 a + 1 )}{9 ^{3}} = \frac{6 a + 18 b}{3} + \frac{- 4 a + 1}{3} =$

$= \frac{6 a + 18 b - 4 a + 1}{3} = \frac{2 a + 18 b + 1}{3} = \frac{2 a}{3} + \frac{18 b}{3} + \frac{1}{3} = \frac{2}{3} a + 6 b + \frac{1}{3}$

$b)$

$\frac{15 + 25 b}{5} - \frac{2 b + 30}{10} = \frac{15 + 25 b}{5} - \frac{2 \cdot b + 2 \cdot 15}{10} = \frac{15 + 25 b}{5} - \frac{2 ( b + 15 )}{10 ^{5}} = \frac{15 + 25 b}{5} - \frac{b + 15}{5} =$

$= \frac{15 + 25 b - ( b + 15 )}{5} = \frac{15 + 25 b - b - 15}{5} = \frac{24 b}{5} = \frac{24}{5} b = 4 \frac{4}{5} b$

$c)$

$12 \cdot \frac{3 x + 12}{6} - 8 \cdot \frac{6 x - 1}{2} = 12^{2} \cdot \frac{3 x + 12}{6} - 8^{4} \cdot \frac{6 x - 1}{2} = 2 \cdot (3 x + 12) - 4 \cdot (6 x - 1) =$

$= 2 \cdot 3 x + 2 \cdot 12 - 4 \cdot 6 x - 4 \cdot (- 1) = \underline{6 x} + \underline{\underline{24}} - \underline{24 x} + \underline{\underline{4}} = - 18 x + 28$

$d)$

$28 \cdot \frac{1 + 7 y}{7} + 6 \cdot \frac{12 - 2 y}{24} = 28^{4} \cdot \frac{1 + 7 y}{7} + 6 \cdot \frac{2 \cdot 6 - 2 \cdot y}{24 ^{4}} = 4 \cdot (1 + 7 y) + \frac{2 ( 6 - y )}{4 ^{2}} =$

$= 4 \cdot 1 + 4 \cdot 7 y + \frac{6 - y}{2} = 4 + 28 y + \frac{6}{2} - \frac{y}{2} = \underline{\underline{4}} + \underline{\frac{56}{2} y} + \underline{\underline{3}} - \underline{\frac{y}{2}} = \frac{55}{2} y + 7 = 27 \frac{1}{2} y + 7$

$e)$

$15 \cdot \frac{x + 16}{30} - 9 \cdot \frac{12 - x}{18} = 15 \cdot \frac{x + 16}{30 ^{2}} - 9 \cdot \frac{12 - x}{18 ^{2}} = \frac{x + 16}{2} - \frac{12 - x}{2} = \frac{x}{2} + \frac{16}{2} - (\frac{12}{2} - \frac{x}{2}) =$

$= \frac{x}{2} + 8 - (6 - \frac{x}{2}) = \underline{\underline{\frac{x}{2}}} + \underline{8} - \underline{6} + \underline{\underline{\frac{x}{2}}} = \frac{2 x}{2} + 2 = x + 2$

$f)$

$13 \cdot \frac{3 a - 9 b}{39} + \frac{12 a + 3 b}{9} = 13 \cdot \frac{3 a - 9 b}{39 ^{3}} + \frac{3 \cdot 4 a + 3 \cdot b}{9} = \frac{3 a - 9 b}{3} + \frac{3 ( 4 a + b )}{9} = \frac{3 a - 9 b}{3} + \frac{3 ( 4 a + b )}{9 ^{3}} =$

$= \frac{3 a - 9 b}{3} + \frac{4 a + b}{3} = \frac{3 a - 9 b + 4 a + b}{3} = \frac{7 a - 8 b}{3} = \frac{7 a}{3} - \frac{8 b}{3} = 2 \frac{1}{3} a - 2 \frac{2}{3} b$

$g)$

$17 \cdot \frac{17 a - 8 b}{34} - 7 \cdot \frac{14 a - 4 b}{2} = 17 \cdot \frac{17 a - 8 b}{34 ^{2}} + \frac{- 7 \cdot 14 a - ( - 7 ) \cdot 4 b}{2} = \frac{17 a - 8 b}{2} + \frac{- 98 a + 28 b}{2} =$

$= \frac{17 a - 8 b - 98 a + 28 b}{2} = \frac{- 81 a + 20 b}{2} = \frac{- 81}{2} a + \frac{20 b}{2} = - 40, 5 a + 10 b$

$h)$

$4 \cdot \frac{40 a + 18 b}{36} - 6 \cdot \frac{9 a - 8 b}{18} = 4 \cdot \frac{40 a + 18 b}{36 ^{9}} - 6 \cdot \frac{9 a - 8 b}{18 ^{3}} = \frac{40 a + 18 b}{9} - \frac{9 a - 8 b}{3} = \frac{40 a + 18 b}{9} - \frac{3 \cdot ( 9 a - 8 b )}{3 \cdot 3} =$

$= \frac{40 a + 18 b}{9} - \frac{27 a - 24 b}{9} = \frac{40 a + 18 b - ( 27 a - 24 b )}{9} = \frac{40 a + 18 b - 27 a + 24 b}{9} = \frac{13 a + 42 b}{9} =$

$= \frac{13 a}{9} + \frac{42 b}{9} = 1 \frac{4}{9} a + 4 \frac{6}{9} b = 1 \frac{4}{9} a + 4 \frac{2}{3} b$

Ewelina Wysopal

Nauczycielka matematyki

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