Question

1.75<4x-x²<4 решите двойное неравенство​

Answer 1

$1,75<4x-x^2<4\; \; \Rightarrow \; \; \; \left \{ {{4x-x^2-1,75>0} \atop {4x-x^2-4<0}} \right. \; \left \{ {{4x^2-16x+7<0} \atop {x^2-4x+4>0}} \right. \\\\\left \{ {{4(x-3,5)(x-0,5)<0} \atop {(x-2)^2>0}} \right. \; \left \{ {{x\in (0,5\, ;\, 3,5)} \atop {x\ne 2}} \right. \; \; \Rightarrow \; \; \; x\in (0,5\, ;\, 2)\cup (2\, ;\, 3,5)$

$P.S.\\\\(x-2)^2\geq 0\; \; pri\; \; x\in (-\infty ,+\infty )\; \; \Rightarrow \; \; \left \{ {{(x-2)^2>0\; \; pri\; \; x\in (-\infty ,2)\cup (2,+\infty )\; ,} \atop {(x-2)^2=0\; \; pri\; \; x=2\; .\qquad \qquad \quad \; }} \right.\; \\\\ili\; \; \; (x-2)^2\geq 0\; \; \; \Rightarrow \; \; \; \left \{ {{(x-2)^2>0\; \; pri\; \; x\ne 2\; ,} \atop {(x-2)^2=0\; \; pri\; \; x=2.}} \right.$