Răspuns:
Explicație pas cu pas:
c) x!/(x-3)! + x!/[x-(x-2)]!(x-2)!=14x
⇒1·2·3·....(x-3)(x-2)(x-1)x/[1·2·3·....(x-3)]+1·2·3·....(x-3)(x-2)(x-1)x/2!·[1·2·3·....(x-3)(x-2)]=14x ⇒(x-2)(x-1)x+(x-1)x/2!=14x ⇒2·(x-2)(x-1)x+(x-1)x=2·14x ⇒x[2(x²-3x+2)+(x-1)]=28x ⇒x(2x²-4x+4+x-1)=28x ⇒x(2x²-3x+3-28)=0 ⇒
Un produs este zero cand oricare din termeni este zero. ⇒Prima solutie este x=0 ∈N dar nu convine
Daca x≠0 ⇒2x²-3x-25=0 ⇒Δ=25+8·25=25+200=225
⇒x₁,₂=[-(-5)±√225]/2·2=(5±15)/4 ⇒
x₁=(5+15)/4=20/4=5 x₁=5 ∈N
x₂=[-(-5)-√225]/2·2=(5-15)/4=-10/4 x₂=-5/2 ∉N
Solutia este x₁=5
d) (x+10)!/ (x+5)!·5!= 5·(x+8)!/(x+5)! ⇒(x+5)!·(x+6)(x+7)(x+8)(x+9)(x+10)/(x+5)!·5!=5·(x+5)!(x+6)(x+7)(x+8)/(x+5)! Deoarece x∈N, (x+5)≠0 si se poate simplifica prin (x+5)! ⇒
⇒(x+6)(x+7)(x+8)(x+9)(x+10)/5!= 5·(x+6)(x+7)(x+8) Deoarece x∈N, putem sa simplificam prin (x+6)(x+7)(x+8) ⇒(x+9)(x+10)/5!=5 ⇒(x+9)(x+10)/2·3·4·5= 5 ⇒(x+9)(x+10)/120=5 ⇒(x+9)(x+10)=120·5 ⇒x²+19x+90=600
⇒x²+19x-510=0 Calculam Δ=b²-4ac=19²-4·1·(-510)=361+2040=2401 ⇒
x₁,₂=( -19±√2401)/2·1 ⇒x₁=(-19+√2401)/2=(-19+49)/2=30/2=15
x₂=(-19-√2401)/2=(-19-49)/2=-68/2=-34
x₁=15 ∈N
x₂=-34 ∉N
Solutia ecuatiei este x₁=15
