Wykonaj działania...- Zadanie 13.32: Matematyka i przykłady jej zastosowań 2. Zakres podstawowy i rozszerzony

Rozwiązanie

a) Dziedzina

$x + 2 = 0 lub x - 2 = 0$

$x = - 2 lub x = 2$

$D = R \ {- 2, 2}$

Wykonujemy działania

$p + r = \frac{x - 2}{x + 2} + \frac{x + 2}{x - 2} = \frac{( x - 2 ) ^{2}}{( x - 2 ) ( x + 2 )} + \frac{( x + 2 ) ^{2}}{( x - 2 ) ( x + 2 )} = \frac{x ^{2} - 4 x + 4 + x ^{2} + 4 x + 4}{x ^{2} - 4} = \frac{2 x ^{2} + 4}{x ^{2} - 4}$

$p - r = \frac{x - 2}{x + 2} - \frac{x + 2}{x - 2} = \frac{( x - 2 ) ^{2}}{( x - 2 ) ( x + 2 )} - \frac{( x + 2 ) ^{2}}{( x - 2 ) ( x + 2 )} = \frac{x ^{2} - 4 x + 4 - x ^{2} - 4 x - 4}{x ^{2} - 4} = \frac{- 8 x}{x ^{2} - 4}$

$p \cdot r = \frac{x - 2}{x + 2} \cdot \frac{x + 2}{x - 2} = 1$

$p : r = \frac{x - 2}{x + 2} : \frac{x + 2}{x - 2} = \frac{x - 2}{x + 2} \cdot \frac{x - 2}{x + 2} = \frac{( x - 2 ) ^{2}}{( x + 2 ) ^{2}}$

b) Dziedzina

$1 - x^{2} = 0 lub (1 - x)^{2} = 0$

$(1 - x) (x + 1) = 0 lub 1 - x = 0$

$x = 1 lub x = - 1$

$D = R \ {- 1, 1}$

Wykonujemy działania

$p + r = \frac{x}{1 - x ^{2}} + \frac{1}{( 1 - x ) ^{2}} = \frac{x}{( 1 - x ) ( 1 + x )} + \frac{1}{( 1 - x ) ^{2}} = \frac{x ( 1 - x )}{( 1 - x ) ^{2} ( 1 + x )} + \frac{1 + x}{( 1 - x ) ^{2} ( 1 + x )} = \frac{x - x ^{2} + 1 + x}{( 1 - x ) ^{2} ( 1 + x )} = \frac{- x ^{2} + 2 x + 1}{( 1 - x ) ^{2} ( 1 + x )}$

$p - r = \frac{x}{1 - x ^{2}} - \frac{1}{( 1 - x ) ^{2}} = \frac{x}{( 1 - x ) ( 1 + x )} - \frac{1}{( 1 - x ) ^{2}} = \frac{x ( 1 - x )}{( 1 - x ) ^{2} ( 1 + x )} - \frac{1 + x}{( 1 - x ) ^{2} ( 1 + x )} = \frac{x - x ^{2} - 1 - x}{( 1 - x ) ^{2} ( 1 + x )} = \frac{- x ^{2} - 1}{( 1 - x ) ^{2} ( 1 + x )}$

$p \cdot r = \frac{x}{1 - x ^{2}} \cdot \frac{1}{( 1 - x ) ^{2}} = \frac{x}{( 1 - x ) ( 1 + x )} \cdot \frac{1}{( 1 - x ) ^{2}} = \frac{x}{( 1 - x ) ^{3} ( 1 + x )}$

$p : r = \frac{x}{1 - x ^{2}} : \frac{1}{( 1 - x ) ^{2}} = \frac{x}{1 - x ^{2}} \cdot \frac{( 1 - x ) ^{2}}{1} = \frac{x}{( 1 - x ) ( 1 + x )} \cdot \frac{( 1 - x ) ^{2}}{1} = \frac{x ( 1 - x )}{1 + x}$

c) Dziedzina

$x^{2} - 3 x + 2 = 0 lub x - 2 = 0 lub x^{2} - 1 = 0$

$x^{2} - x - 2 x + 2 = 0 lub x = 2 lub (x - 1) (x + 1) = 0$

$x (x - 1) - 2 (x - 1) = 0 lub x = 2 lub x - 1 = 0 lub x + 1 = 0$

$(x - 1) (x - 2) = 0 lub x = 2 lub x = 1 lub x = - 1$

Zatem

$x = 2 lub x = 1 lub x = - 1$

$D = R \ {- 1, 1, 2}$

Wykonujemy działania

$p + r = \frac{x + 1}{x ^{2} - 3 x + 2} + \frac{x - 2}{x ^{2} - 1} = \frac{x + 1}{( x - 1 ) ( x - 2 )} + \frac{x - 2}{( x - 1 ) ( x + 1 )} = \frac{( x + 1 ) ^{2}}{( x - 1 ) ( x - 2 ) ( x + 1 )} + \frac{( x - 2 ) ^{2}}{( x - 1 ) ( x - 2 ) ( x + 1 )} = \frac{x ^{2} + 2 x + 1 + x ^{2} - 4 x + 4}{( x - 1 ) ( x - 2 ) ( x + 1 )} = \frac{2 x ^{2} - 2 x + 5}{( x - 1 ) ( x - 2 ) ( x + 1 )}$

$p - r = \frac{x + 1}{x ^{2} - 3 x + 2} - \frac{x - 2}{x ^{2} - 1} = \frac{x + 1}{( x - 1 ) ( x - 2 )} - \frac{x - 2}{( x - 1 ) ( x + 1 )} = \frac{( x + 1 ) ^{2}}{( x - 1 ) ( x - 2 ) ( x + 1 )} - \frac{( x - 2 ) ^{2}}{( x - 1 ) ( x - 2 ) ( x + 1 )} = \frac{x ^{2} + 2 x + 1 - x ^{2} + 4 x - 4}{( x - 1 ) ( x - 2 ) ( x + 1 )} = \frac{6 x - 3}{( x - 1 ) ( x - 2 ) ( x + 1 )}$

$p \cdot r = \frac{x + 1}{x ^{2} - 3 x + 2} \cdot \frac{x - 2}{x ^{2} - 1} = \frac{x + 1}{( x - 1 ) ( x - 2 )} \cdot \frac{x - 2}{( x - 1 ) ( x + 1 )} = \frac{1}{( x - 1 ) ^{2}}$

$p : r = \frac{x + 1}{x ^{2} - 3 x + 2} : \frac{x - 2}{x ^{2} - 1} = \frac{x + 1}{( x - 1 ) ( x - 2 )} \cdot \frac{( x - 1 ) ( x + 1 )}{x - 2} = \frac{( x + 1 ) ^{2}}{( x - 2 ) ^{2}}$