√2-√x<=x
C.E: x>=0 => x€[0,+infinit]
Ridicand la patrat avem:
4-2√2x+x<=x^2
-2√2x<=x^2-x-4
C.E: x^2-x-4>=0
Egalam cu 0:
x^2-x-4=0
∆=1+16
∆=17
x1=(1+√17)/2 => x ~=2.56
x2=(1-√17)/2 => x~=-1.56
Acum cu tabel de semne vom obtine:
x€(-infinit,(1-√17)/2) U ((1+√17)/2,+infinit)
dar x€(0,+infinit)
Din intersectia celor doua rezulta:
x€((1+√17)/2,+infinit)
Revenind la ecuatie:
-2√2x<=x^2-x-4 /^2
8x<=x^4+x^2+16-2x^3-8x^2+8x
x^4-2x^3-7x^2<=16
x^2(x^2-2x-7)<=16
Egalam:
x^2(x^2-2x-7)=16
/// I. x^2=16 => x=4
/// II. x^2-2x-7=1
x^2-2x-8=0
∆=4+32
∆=36
x1'=(2+6)/2 => x1'=4
x2'=(2-6)/2 => x2'=-2
Din nou tabel de semne(noua ne trebuia mai mic sau egal):
=> x€[-2,4]
dar x€((1+√17)/2,+infinit)
Intersectand cele doua intervale obtinem intervalul final:
[(1+√17)/2,4] #
Bafta!
