Question

При каких значениях х значение дроби равно $\frac{16 - {x}^{2} }{x - 4}$

$\frac{ {x}^{2} - 2x - 35 }{x - 7}$

Answer 1

1.  

$\frac{16-x^{2} }{x-4} =0 < = \left \{ {{16-x^{2} =0} \atop {x-4\neq 0}} \right. < = \left \{ {{x^{2}-16 =0} \atop {x\neq 4}} \right. < = \left \{ {{(x-4)(x+4)=0} \atop {x\neq 0}} \right. < =$

$< = \left \{ {{x=-4;x=4} \atop {x\neq 4}} \right. < = x=-4$

ответ:   при $x=-4$

2.

$\frac{x^{2}-2x-35 }{x-7} =0 < = \left \{ {{x^{2} -2x-35=0} \atop {x-7\neq 0}} \right. < = \left \{ {{(x-7)(x+5)=0} \atop {x\neq 7}} \right. < = \left \{ {{x=-5;x=7} \atop {x\neq 7}} \right. < =$

$< = x=-5$

ответ:   при $x=-5$