Question

Срочно помогите сделать, плачу 15 баллов!!!!!

Answer 1

$\frac{x^{2}-7x+10 }{2-x^{2}+x} \geq 0$ | : (-1)

$\frac{x^{2}-7x+10 }{x^{2}-x-2} \leq 0$

Дробь может быть меньше нуля в двух случаях:

$\left[\begin{array}{ccc}\left \{ {{x^{2}-7x+10 \leq 0 } \atop {x^{2} -x-2>0}} \right.\\\left \{ {{x^{2} -x-2<0} \atop {x^{2}-7x+10 \geq 0 }} \right.\\\end{array}$

x² - 7x + 10 = 0

D = (-7)² - 4*10 = 49 - 40 = 9 = 3²

$x_{1} = \frac{7+3}{2*1} = \frac{10}{2} = 5$

$x_{2} = \frac{7-3}{2*1} = \frac{4}{2} = 2$

x² - x - 2 = 0

D = (-1)² - 4*(-2) = 1 + 8 = 9 = 3²

$x_{1} = \frac{1+3}{2*1} = \frac{4}{2} = 2$

$x_{2} = \frac{1-3}{2*1} = - \frac{2}{2} = -1$

Т.к. на 0 делить нельзя ⇒ х ≠ -1 и х ≠ 2

$\left[\begin{array}{ccc}\left \{ {{2\leq x \leq 5 } \atop {\left[\begin{array}{ccc}x<-1\\x>2\\\end{array}}} \right.\\\left \{ {{-1<x<2 } \atop {\left[\begin{array}{ccc}x\leq 2\\x\geq 5\\\end{array}}} \right.\\\end{array}$

$\left[\begin{array}{ccc}\left \{ {{2\leq x\leq 5 } } \right.\\\left \{ {{-1<x<2 } } \right.\\\end{array}$

Ответ: х∈(-1;2)∪(2;5]