Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]a)~~\dfrac{7}{x-3} -\dfrac{2}{x+3} -\dfrac{12}{x^{2} -9} =\dfrac{7}{x-3} -\dfrac{2}{x+3} -\dfrac{12}{(x+3)(x-3)}=\dfrac{7(x+3)-2(x-3)-12}{(x+3)(x-3)}=\dfrac{7x+21-2x+6-12}{(x+3)(x-3)}=\dfrac{5x+15}{(x+3)(x-3)}=\dfrac{5(x+3)}{(x+3)(x-3)}=\dfrac{5}{(x-3)}\\\\zal.\\x-3\neq 0~~\Rightarrow ~~x\neq 3\\x+3\neq 0~~\Rightarrow ~~x\neq -3\\D=R-\{ -3,3\}[/tex]
[tex]b)~~\dfrac{5}{x+4} -\dfrac{2}{x-4} +\dfrac{40}{x^{2} -16} =\dfrac{5}{x+4} -\dfrac{2}{x-4} +\dfrac{40}{(x+4)(x-4)} =\dfrac{5(x-4)-2(x+4)+40}{(x+4)(x-4)} =\dfrac{5x-20-2x-8+40}{(x+4)(x-4)} =\dfrac{3x+12}{(x+4)(x-4)} =\dfrac{3(x+4)}{(x+4)(x-4)} =\dfrac{3}{(x-4)} \\\\zal.\\x+4\neq 0~~\Rightarrow ~~x\neq -4\\x-4\neq 0~~\Rightarrow ~~x\neq 4\\D=R-\{ -4,4\}[/tex]
Korzystam ze wzoru skróconego mnożenia:
a² - b² = ( a+b)(a-b)
