Решите уравнения: а)√x^2-3x-18=3-x . (Пояснение для уравнения а : Вся левая часть до ...

а) $\sqrt{x^{2}-3 x-18}=3-x$

ОДЗ: $\left \{ {{x^{2}-3x-18\geq 0} \atop {3-x\geq0 }} \right.$ " alt="=>" align="absmiddle" class="latex-formula"> $\left \{ {{(x+3)(x-6)\geq 0} \atop {x\leq 3 }} \right.$ " alt="=>" align="absmiddle" class="latex-formula"> $\left \{ {{x\leq -3;x\geq 6} \atop {x\leq 3}} \right.$ " alt="=>" align="absmiddle" class="latex-formula"> $x\leq -3$

$(\sqrt{x^{2}-3 x-18})^2=(3-x)^2$

$x^{2}-3 x-18=(3-x)^2$

$x^{2}-3 x-18=9-6x+x^{2}$

$x^{2}-3 x-18-9+6x-x^{2}=0$

$3 x-27=0$

$3 x=27$

$x=27:3$

$x=9$ не удовлетворяет ОДЗ

Ответ: {∅}

б) $log_4(4+7x)-log_4(1+5x)=1$

ОДЗ: 0} \atop {1+5x>0}} \right. =>\left \{ {{x>-\frac{4}{7} } \atop {x>-\frac{1}{5} }} \right.=>x>-\frac{1}{5}" alt="\left \{ {{4+7x>0} \atop {1+5x>0}} \right. =>\left \{ {{x>-\frac{4}{7} } \atop {x>-\frac{1}{5} }} \right.=>x>-\frac{1}{5}" align="absmiddle" class="latex-formula">