Question

(3-2√2)ˣ+(3+2√2)ˣ=6
Rezolvati ecuatia

Answer 1

"se observa" ca 3-2√2=1/(3+2√2)
intr-adevar
1/(3+2√2)=(3-2√2)/(9-8)=(3-2√2)

atunci notand (3+2√2)^x = t , avem ecuatia

1/t+t=6| *t
1+t²=6t
t²-6t+1=0
t1,2=(6+/-√(36-4))/2=(6+/-4√2)/2=(3+/-2√2)

t1=3+2√2=(3+2√2) ^x⇒x=1
t2=3-2√2=(3+2√2)^x⇒x=-1

s= {-1;1}
grea!!!!!

Answer 2

1/3+2√3 prin raționalizare 3-2√2/(3-√2)(3×2√3)=3-2√2/9-8=3-2√2
notanx
(3+2√2)^x=t
1/t+t=6
1+t^2-6t=0
∆=36-4=32
t1=6-4√2/2
t1=3-2√2
(3+2√2)^x=1/3+2√2=(3+2√2)^-1
x=-1
t2=6+4√2/2
t2=3+2√2
(3+2√2)^x=(3+2√2)^1
x=1

sau o alta rezolvarea

2√2=2ab|:2
a=1
b=√2
(3-2√2)=(1-√2)^2
(3+2√2)=(1+√2)^2
inlocuind in ec inițială
(1-√2)^2x+(1+√2)^2x=6
1/1+√2 prin raționalizare 1-√2/(1-√2)(1+√2)=1-√2
(1/1+√2)^2x+(1+√2)^2x=6
notăm
(1+√2)^2x=t
1/t+t-6=0
t^2-6t+1=0
t1=(3-2√2)=(1-√2)^2
2x=-2
x=-1
t2=(3+2√2)=(1+√2)^2
2x=2
x=1