Question

Помогите пожалуйста с 3и 4 заданием.

Answer 1

Ответ:

3) x≤3 x<1 4) x1=3 x2=2 ............

Answer 2

$3)\; \; \frac{x-3}{log_5x}\leq 0\; \; ,\; \; ODZ:\; \left \{ {{log_5x\ne 0} \atop {x>0}} \right.\; \; \left \{ {{x\ne 1} \atop {x>0}} \right.\\\\\left \{ {{x-3\leq 0} \atop {log_5x>0}} \right.\; \; ili\; \; \; \left \{ {{x-3\geq 0} \atop {log_5x<0}} \right.\\\\\left \{ {{x\leq 3} \atop {x>1}} \right.\; \; \; \; \; \; ili\quad \; \; \left \{ {{x\geq 3} \atop {x<1}} \right.\\\\x\in (1,3\, ]\; \; \; ili\; \; \; \; x\in \varnothing \\\\Otvet:\; \; x\in (1,3\, ]\; .$

$4)\; \; \frac{6^{x^2}}{3^2}=\frac{2^2}{6^{8-5x}}\\\\6^{x^2}\cdot 6^{8-5x}=2^2\cdot 3^2\; \; ,\qquad 2^2\cdot 3^2=(2\cdot 3)^2=6^2\\\\6^{x^2+8-5x}=6^2\\\\x^2+8-5x=2\\\\x^2-5x+6=0\; \; ,\; \; x_1=2\; ,\; x_2=3\; \; (teorema\; Vieta)\\\\Otvet:\; \; x=2\; ,\; \; x=3\; .$