[tex] \sqrt{x+2}+ \sqrt{3-x} =3 [/tex]
Cond:
x+2≥0=> x≥-2 =>x ∈[-2;+∞)
3-x≥0 <=> -x≥-3 <=> x≤3 =>x∈(-∞;3 ]
=> x∈[-2;3]
Ridicam la patrat si obt:
x+2 +3-x +2([tex] \sqrt{(x+2)(3-x)} [/tex]=9
<=> 5+2[tex] \sqrt{(-x^2+x+6)} [/tex]=9<=> 2[tex] \sqrt{(-x^2+x+6)} [/tex]=4 |2 <=> 4(-x²+x+6)= 16 |:4 <=> -x²+x+6= 4 |(-1)<=> x²-x-2=0
Δ=1+8= 9 => √Δ=3
x1=2 si x2= -1
S={-1; 2}
