Question

помогите плииииз прошу

Answer 1

$3\sqrt{1\frac{9}{16}}-1=3\sqrt{\frac{25}{16}}-1=3\cdot \frac{5}{4}-1=\frac{15-4}{4}=\frac{11}{4}\\\\\sqrt{2,5}\cdot \sqrt{10}=\sqrt{2,5\cdot 10}=\sqrt{25}=5\\\\\frac{\sqrt{0,72}}{\sqrt8}=\sqrt{\frac{72}{100\cdot 8}}=\sqrt{\frac{9}{100}}=\frac{3}{10}=0,3\\\\\sqrt{3^4\cdot 2^6}=3^2\cdot 2^3=9\cdot 8=72\\\\2)\; \; (x-4)(x+7)\ \textless \ 0;\; \; \; +++(-7)---(4)+++\\\\x\in (-7,4)\\\\3)\frac{x-8}{x+3}\ \textgreater \ 0;\; \; \; \; +++(-3)---(8)+++\\\\x\in (-\infty,-3)U(8,+\infty)\\\\4)\; \; x^3-49x\ \textgreater \ 0\\\\x(x-7)(x+7)\ \textgreater \ 0$

$---(-7)+++(0)---(7)+++\\\\x\in (-7,0)U(7,+\infty)\\\\5)\; \; 6x^2-7x-24\ \textless \ 0\\\\D=625,\; x_1=\frac{7-25}{12}=-\frac{18}{12}$

$x_1=-\frac{3}{2}=-1,5$

$x_2=\frac{7+25}{12}=\frac{32}{12}=\frac{8}{3}\\\\+++(-1,5)---(\frac{8}{3})+++\\\\x\in (-1,5\, ;\, \frac{8}{3})$