Question

Buna Ziua!As avea nevoie de ajutorul vostru! Va dau coroana si puncte numai daca rezolvati urmatoarea inegalitate
${ (\sqrt{5 + 2 \sqrt{6} }) }^{x} + { (\sqrt{5 - 2 \sqrt{6} }) }^{x} < 10$
Am folosit formula radicalilor compusi, dar in rest n-am nici o idee ce trebuie sa fac.
$\sqrt{5 + 2 \sqrt{6} } = \sqrt{3} + \sqrt{2}$
$\sqrt{5 - 2 \sqrt{6} } = \sqrt{3} - \sqrt{2}$

Answer 1

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Answer 2

(√3+√2)^x+(√3-√2)^x=10
fie √3+√2=a>1>0
atunci 1/a= 1/(√3+√2)=(√3-√2)/(3-2)=√3-√2

ecuatia noastra devine
a^x+1/a^x=10
fie a^x=y
y+1/y=10
y²-10y+1=0
y1,2=(10+-√96)/2=(10+-4√6)/2=5+-2√6

a^x=5+2√6
(√3+√2)^x=5+2√6
se observa ca x=2 verifica relatia
intr-adevar 3+2√2*√3+2=5+2√6
a>1,  a^x injectiva (crescatoare), deci x=2 solutie unica a ecuatiei exponentiale (√3+√2)^x=5+2√6

a^x=5-2√6=1/(5+2√6)...intr-adevar1/(5+2√6)=(25-24)=(5-2√6)/1
atunci
(√3+√2)^x=1/(√3+√2)²=(√3+√2)^(-2)
egalam exponentii
x=-2
(√3+√2)^x injectiva pe R, x=-2, solutie unica


deci S={-2;2}