Proszę o szybkie zrobienie.

[tex]-x^2+5x\geq 0\quad\text{i}\quad 2x-4>0\\\\x^2-5x\leq0\quad\text{i}\quad 2x>4\\\\x(x-5)\leq0\quad\text{i}\quad x>2\\\\x\in\langle0,5\rangle\quad\text{i}\quad x\in(2,+\infty)[/tex]

[tex]\boxed{\text{D}=(2,5\rangle}[/tex]

Zadanie 3

[tex]x^2-9>0\quad\text{i}\quad 3-\dfrac{x}{7}>0\\\\(x+3)(x-3)>0\quad\text{i}\quad x<21\\\\x\in(-\infty,-3)\cup(3,+\infty)\quad\text{i}\quad x\in(-\infty,21)[/tex]

[tex]\boxed{\text{D}=(-\infty,-3)\cup(3,21)}[/tex]

Unicorn05 Unicorn05 · Answer 2 · 2021-11-06T10:09:12+01:00

1)

[tex]f(x)=\dfrac{\sqrt{-3x+6}+8x}{x+4}\\\\D:\\{}\qquad -3x+6\ge0\qquad\ \ \wedge\qquad x+4\ne0\\\\ {}\quad -3x\ge-6\quad/:(-3)\qquad \wedge\qquad x\ne-4\\\\{}\qquad\qquad x\le2\qquad\ \ \wedge\qquad\ \ x\ne-4\\\\D=(-\infty\,;-4)\cup(-4\,;2\big>[/tex]

2)

[tex]f(x)=\sqrt{-x^2+5x}+\log_3(2x-4)\\\\D:\\{}\qquad-x^2+5x\ge0\qquad\ \wedge\qquad 2x-4>0\\\\ {}\qquad-x(x-5)\ge0\qquad\ \wedge\qquad 2x>4\qquad/:2\\{}\qquad^{x_1\,=\,0\,,\ \ x_2\,=\,5}\\{}\qquad\quad\ x\in\big<0\,;\,5\big>\qquad\ \wedge\qquad\ \ x>2\\\\D=(2\,;\, 5\big>[/tex]

3)

[tex]f(x)=\dfrac{4x^2+\sqrt3}{\sqrt{x^2-9}}-ln(3-\frac x7)\\\\D:\\{}\qquad x^2-9>0\qquad\quad \wedge\qquad 3-\frac x7>0\\\\ {}\quad(x+3)(x-3)>0\qquad\ \wedge\quad -\frac17x>-3\qquad/\cdot(-7)\\{}\quad ^{x_1\,=\,-3\,,\ x_2\,=\,3}\\ x\in(-\infty\,;-3)\cup(3\,;\,\infty)\qquad\ \wedge\quad x<21\\\\D=(-\infty\,;-3)\cup(3\,;\,21)[/tex]