Question

$\frac{ {x}^{2} - 4 }{x + 2 } + \frac{ {x}^{2} - 3x + 2 \: \: \: }{x - 1} para \: x = 4$
me ajudem porfavor​

Answer 1

Se x=4 basta substituir os valores:

$\frac{ {x}^{2} - 4 }{x + 2 } + \frac{ {x}^{2} - 3x + 2 \: \: \: }{x - 1} \\\\\frac{ {4}^{2} - 4 }{4 + 2 } + \frac{ {4}^{2} - 3(4) + 2 \: \: \: }{4 - 1}\\ \\\frac{16 - 4}{6} + \frac{16-12+2}{3}\\\\\frac{12}{6}+\frac{2}{3}\\\\\frac{12 + 4}{6} = \frac{16}{6} = \frac{8}{3}$