Wykonaj redukcję wyrazów podobnych i oblicz wartość liczbową - Zadanie 4: Matematyka 7

Rozwiązanie

a) $\underline{2 a} - \underline{\underline{3 b}} + \underline{\underline{\underline{4 c^{2}}}} - \underline{7 a} + \underline{\underline{12 b}} - \underline{\underline{\underline{3 \frac{1}{2} c^{2}}}} = - 5 a + 9 b + \frac{1}{2} c^{2} =$

$= - 5 \cdot (- 3) + 9^{3} \cdot (- \frac{1}{6 ^{2}}) + \frac{1}{2} \cdot 2^{2} = 15 - \frac{3}{2} + \frac{1}{2} \cdot 4 = 15 - 1 \frac{1}{2} + 2 = 15 \frac{1}{2}$

b) $\underline{- 3 a b c} + \underline{\underline{2 a^{2} b^{2}}} - 3 a b - \underline{\underline{a^{2} b^{2}}} + \underline{4 a b c} = a b c + a^{2} b^{2} - 3 a b =$

$= - 3^{1} \cdot (- \frac{1}{6 ^{2}}) \cdot 2 + (- 3)^{2} \cdot (- \frac{1}{6})^{2} - 3 \cdot (- 3^{1}) \cdot (- \frac{1}{6 ^{2}}) = 1 + 9^{1} \cdot \frac{1}{36 ^{4}} - \frac{3}{2} =$

$= 1 + \frac{1}{4} - \frac{6}{4} = 1 - \frac{5}{4} = 1 - 1 \frac{1}{4} = - \frac{1}{4}$ $= 1 + \frac{1}{4} - \frac{6}{4} = 1 - \frac{5}{4} = 1 - 1 \frac{1}{4} = - \frac{1}{4}$

c) $\underline{7 a c} - \underline{\underline{4 a b}} + \underline{\underline{\underline{8 b c}}} - \underline{6 a c} + \underline{\underline{12 a b}} - \underline{\underline{\underline{10 b c}}} = a c + 8 a b - 2 b c =$

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

d) $\underline{- 2 a^{3} b^{2} c} + 2 a b^{2} c^{3} - 3 a^{2} b^{2} c^{2} + \underline{4 a^{3} b^{2} c} = 2 a^{3} b^{2} c + 2 a b^{2} c^{3} - 3 a^{2} b^{2} c^{2} =$

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

$= 2 \cdot (- 27^{3}) \cdot \frac{1}{36 ^{4}} \cdot 2 - 6^{1} \cdot \frac{1}{36 ^{6}} \cdot 8 - 3 \cdot 9^{1} \cdot \frac{1}{36 ^{4}} \cdot 4 =$

$= - 6 \cdot \frac{2}{4} - \frac{8}{6} - 3 \cdot \frac{4}{4} = - 6 \cdot \frac{1}{2} - \frac{4}{3} - 3 \cdot 1 = - 3 - 1 \frac{1}{3} - 3 = - 7 \frac{1}{3}$

e) $\underline{0, 6 b^{2} c} - \underline{\underline{0, 2 a^{2} b}} + \underline{1, 4 c b^{2}} - \underline{\underline{0, 8 b a^{2}}} - \frac{1}{9} = 2 b^{2} c - a^{2} b - \frac{1}{9} =$

$= 2 \cdot (- \frac{1}{6})^{2} \cdot 2 - (- 3)^{2} \cdot (- \frac{1}{6}) - \frac{1}{9} = 2 \cdot \frac{1}{36} \cdot 2 - 9 \cdot (- \frac{1}{6}) - \frac{1}{9} =$

$= \frac{4}{36} + \frac{9}{6} - \frac{1}{9} = \frac{1}{9} + \frac{3}{2} - \frac{1}{9} = \frac{3}{2} = 1 \frac{1}{2}$

f) $\underline{- 3 a b c^{3}} + \underline{\underline{2 b a^{2} c}} - \underline{6 c^{3} a b} - \underline{\underline{4 a^{2} b c}} - b a c^{2} = - 9 a b c^{3} - 2 a^{2} b c - a b c^{2} =$

$= - 9 \cdot (- 3^{1}) \cdot (- \frac{1}{6 ^{2}}) \cdot 2^{3} - 2^{1} \cdot (- 3)^{2} \cdot (- \frac{1}{6 ^{3}}) \cdot 2 - (- 3^{1}) \cdot (- \frac{1}{6 ^{2}}) \cdot 2^{2} =$