Неравенство и сис-ма уровнений помогите плз

Question 1

Question 2

$1)\; 9^{\sqrt{x}}=(3^2)^{\sqrt{x}}=(3^{\sqrt{x}})^2=t^2,\; t=3^{\sqrt{x}}$

$t^2-10t+9<0\\\\t_1=1,t_2=9\; (teor.\; Vieta)$

$+ + + + (1)- - - - -(9)+ + + +$
$1<t<9$

3^0}\atop {3^{\sqrt{x}}<3^2}} \right. \; \left \{ {{\sqrt{x}>0} \atop {\sqrt{x}<2}} \right. \; \left \{ {{x>0} \atop {x<4}} \right. \\\\x\in (0,4)" alt="\left \{{{3^{\sqrt{x}}>3^0}\atop {3^{\sqrt{x}}<3^2}} \right. \; \left \{ {{\sqrt{x}>0} \atop {\sqrt{x}<2}} \right. \; \left \{ {{x>0} \atop {x<4}} \right. \\\\x\in (0,4)" align="absmiddle" class="latex-formula">

$2)\left \{ {{x^2y^2=2x^4-1} \atop {x^2y^2=3x^4-2}} \right.$

$2x^4-1=3x^4-2\\\\x^4=1\; \to \; x^2=1\; \to \; x=\pm 1\\\\ a)\; \left \{ {{x=-1} \atop {2=y^2+1}} \right. \; \left \{ {{x=-1} \atop {y^2=1}} \right. \; \left \{ {{x=-1} \atop {y=\pm 1}} \right. \\\\b)\; \left \{ {{x=1} \atop {2=y^2+1}} \right. \; \left \{ {{x=1} \atop {y=\pm 1}} \right. \\\\Otvet:\; (-1,-1),\; (-1,1),\; (1,-1),\; (1,1).$ <9,\;>

NNNLLL54_zn · Answer 1 · 2018-03-20T02:27:24+0000

$1)\; 9^{\sqrt{x}}=(3^2)^{\sqrt{x}}=(3^{\sqrt{x}})^2=t^2,\; t=3^{\sqrt{x}}$

$t^2-10t+9<0\\\\t_1=1,t_2=9\; (teor.\; Vieta)$

$+ + + + (1)- - - - -(9)+ + + +$
$1<t<9$

3^0}\atop {3^{\sqrt{x}}<3^2}} \right. \; \left \{ {{\sqrt{x}>0} \atop {\sqrt{x}<2}} \right. \; \left \{ {{x>0} \atop {x<4}} \right. \\\\x\in (0,4)" alt="\left \{{{3^{\sqrt{x}}>3^0}\atop {3^{\sqrt{x}}<3^2}} \right. \; \left \{ {{\sqrt{x}>0} \atop {\sqrt{x}<2}} \right. \; \left \{ {{x>0} \atop {x<4}} \right. \\\\x\in (0,4)" align="absmiddle" class="latex-formula">

$2)\left \{ {{x^2y^2=2x^4-1} \atop {x^2y^2=3x^4-2}} \right.$

$2x^4-1=3x^4-2\\\\x^4=1\; \to \; x^2=1\; \to \; x=\pm 1\\\\ a)\; \left \{ {{x=-1} \atop {2=y^2+1}} \right. \; \left \{ {{x=-1} \atop {y^2=1}} \right. \; \left \{ {{x=-1} \atop {y=\pm 1}} \right. \\\\b)\; \left \{ {{x=1} \atop {2=y^2+1}} \right. \; \left \{ {{x=1} \atop {y=\pm 1}} \right. \\\\Otvet:\; (-1,-1),\; (-1,1),\; (1,-1),\; (1,1).$ <9,\;>