Question

ДАМ 20 БАЛЛОВ решите неравенство(любым способом)$x^{2} +x-6\leq 0\\8x-x^{2} -16\ \textgreater \ 0\\5x^{2}+12\geq 6x\\16-x^{2} \ \textless \ 0$

Answer 1

1)

$x^2+x-6\leq 0\\x^2+3x-2x-6\leq0\\x(x+3)-2(x+3)\leq0\\(x+3)(x-2)\leq0$

x ∈ [-3;2]

2)

$8x-x^{2} -16>0\\-(x^2-8x+16)>0\\-(x-4)^2>0\\(x-4)^2<0$

x ∈ ∅

3)

$5x^2+12\geq 6x\\5x^2+12- 6x\geq0\\5x^2+12- 6x=0$

x ∉ R

$5x^2-6x+12\geq0, \to a =5$

x ∈ R

4)

$16-x^2<0\\-x^2=-16\\x^2=16\\x_1 = -4\\x_2= 4$