Question

СРОЧНО !!
сколько целых чисел удовлетворяют неравенству?
$x^{4} - 8x^{2} +7 leq 0$

Answer 1

$x^2-8x+7 leq 0\\t=x^2,; t^2-8t+7 leq 0\\t_1=1,; t_2=7; (teor.; Vieta)\\+++[1]---[7]+++\\ left { {{t geq 1} atop {t leq 7}} right. \\x^2 geq 1; to ; x^2-1 geq 0,; ; (x-1)(x+1) geq 0\\ +++[-1]---[1]+++\\xin (-infty,-1]U[1,+infty)$

$\x^2 leq 7; ; to ; ; x^2-7 leq 0,; ; (x-sqrt7)(x+sqrt7) leq 0\\+++[-sqrt7]---[sqrt7]+++\\xin [-sqrt7,sqrt7]\\\ left { {{xin (-infty,-1]U[1,+infty)} atop {xin [-sqrt7,sqrt7]}} right. ; ; to ; ; xin [-sqrt7,-1]U[1,sqrt7]\\Celue:; x=-2,-1,1,2.$