Question

Решите пожалуйста С1 с обеих вариантов

Answer 1

$1)\; \; \frac{x^2-3x+m}{x+a}=x-5\; \; \; \to \; \; \; x^2-3x+m=(x+a)(x-5)\\\\x^2-3x+m=x^2+(a-5)x-5a\\\\(a-5+3)\cdot x+(-5a-m)=0\; \; \Rightarrow \; \; \left \{ {{a-5+3=0} \atop {-5a-m=0}} \right.\; \; \left \{ {{a=2\quad \; } \atop {m=-5a}} \right. \; \left \{ {{a=2\; \; \; \; \; } \atop {m=-10}} \right. \\\\\\Proverka:\; \; a=2\; ,\; m=-10:\; \; \frac{x^2-3x-10}{x+2}=\frac{(x+2)(x-5)}{x+2}=x-5$

$2)\; \; \frac{x^2+2x+a}{x+b}=x+5\; \; \; \to \; \; \; x^2+2x+a=(x+b)(x+5)\\\\x^2+2x+a=x^2+(5+b)x+5b\\\\(5+b-2)\cdot x+(5b-a)=0\; \; \Rightarrow \; \; \left \{ {{5+b-2=0} \atop {5b-a=0}} \right.\; \; \left \{ {{b=-3\; } \atop {a=5b}} \right. \; \left \{ {{b=-3\; } \atop {a=-15}} \right. \\\\\\Proverka:\; \; a=-15\; ,\; b=-3:\; \; \frac{x^2+2x-15}{x-3}=\frac{(x-3)(x+5)}{x-3}=x+5$