Sprowadź do najprostszej...- Zadanie 1.27: Matura z Matematyki 2018 - ... Andrzej Kiełbasa. Poziom podstawowy i rozszerzony cz. 1

Rozwiązanie

$a) \frac{8 x}{x ^{2} - 8 x} = \frac{8 x}{x \cdot ( x - 8 )} = \frac{8}{x - 8}$

$b) \frac{x ^{3} + 3 x ^{2} - 2 x - 6}{x ^{2} - 2} = \frac{x ^{3} - 2 x + 3 x ^{2} - 6}{x ^{2} - 2} = \frac{x ( x ^{2} - 2 ) + 3 ( x ^{2} - 2 )}{x ^{2} - 2} = \frac{( x + 3 ) ( x ^{2} - 2 )}{x ^{2} - 2} = x + 3$

$c) \frac{k - 1}{k ^{2} + k} \cdot \frac{2 k}{k ^{2} - 1} = \frac{k - 1}{k \cdot ( k + 1 )} \cdot \frac{2 k}{( k - 1 ) ( k + 1 )} = \frac{1}{k + 1} \cdot \frac{2}{k + 1} = \frac{2}{( k + 1 ) ^{2}}$

$d) \frac{2 x}{x + 5} + \frac{10}{x + 5} = \frac{2 x + 10}{x + 5} = \frac{2 ( x + 5 )}{x + 5} = 2$

$e) \frac{8 - 3 x}{x + 2} + 3 = \frac{8 - 3 x}{x + 2} + \frac{3 \cdot ( x + 2 )}{x + 2} = \frac{8 - 3 x}{x + 2} + \frac{3 x + 6}{x + 2} = \frac{8 - 3 x + 3 x + 6}{x + 2} = \frac{14}{x + 2}$

$f) \frac{1}{x + 2} + \frac{1}{x - 2} = \frac{x - 2 + x + 2}{( x + 2 ) ( x - 2 )} = \frac{2 x}{x ^{2} - 4}$

$g) \frac{1}{x - 4} - \frac{1}{x + 4} = \frac{x + 4 - ( x - 4 )}{( x - 4 ) ( x + 4 )} = \frac{8}{x ^{2} - 16}$

$h) \frac{x + 8}{x} - \frac{x + 5}{x - 3} = \frac{( x - 3 ) ( x + 8 ) - x ( x + 5 )}{x \cdot ( x - 3 )} = \frac{x ^{2} + 8 x - 3 x - 24 - x ^{2} - 5 x}{x \cdot ( x - 3 )} = \frac{- 24}{x ^{2} - 3 x}$

$i) \frac{a}{a - b} + \frac{b}{b - a} = \frac{a ( b - a ) + b ( a - b )}{( a - b ) ( b - a )} = \frac{a b - a ^{2} + b a - b ^{2}}{( a - b ) ( b - a )} = \frac{- a ^{2} + 2 a b - b ^{2}}{( b - a ) ( a - b )} = \frac{- ( a ^{2} - 2 a b + b ^{2} )}{- ( a - b ) ( a - b )} = \frac{- ( a - b ) ^{2}}{- ( a - b ) ^{2}} = 1$

$j) \frac{x}{x ^{2} - 9} - \frac{1}{x + 3} = \frac{x}{( x - 3 ) ( x + 3 )} - \frac{x - 3}{( x + 3 ) ( x - 3 )} = \frac{x - ( x - 3 )}{( x + 3 ) ( x - 3 )} = \frac{x - x + 3}{( x + 3 ) ( x - 3 )} = \frac{3}{( x + 3 ) ( x - 3 )} = \frac{3}{x ^{2} - 9}$

Aleksandra Filipowska

Nauczycielka matematyki