(1/4)(3x+1)<(1/6)(5x+1) (3/2)(3x+1)<5x+1 (9/2)x+3/2<5x+1 (9/2)x-5x<1-3 (-1/2)x<-2 (1/2)x>2 x>4 On en déduis que S=]4;+∞[
2x-1/x=2-((3/2)x-1) 2x²-1=2x-x((3/2)x-1) 2x²-1=2x-(3/2)x²+x 2x²-1+(3/2)x²-3x=0 (7/2)x²-3x-1=0 Δ=b²-4ac=(-3)²-4(7/2)(-1)=9+14=23 On a donc 2 racines réelles: x(1)=(-b-√Δ)/2a=(3-√23)/7 x(2)=(-b+√Δ)/2a=(3+√23)/7 On en déduis alors: S={(3-√23)/7;(3+√23)/7}
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Bonjour,(1/4)(3x+1)<(1/6)(5x+1)
(3/2)(3x+1)<5x+1
(9/2)x+3/2<5x+1
(9/2)x-5x<1-3
(-1/2)x<-2
(1/2)x>2
x>4 On en déduis que S=]4;+∞[
2x-1/x=2-((3/2)x-1)
2x²-1=2x-x((3/2)x-1)
2x²-1=2x-(3/2)x²+x
2x²-1+(3/2)x²-3x=0
(7/2)x²-3x-1=0
Δ=b²-4ac=(-3)²-4(7/2)(-1)=9+14=23
On a donc 2 racines réelles:
x(1)=(-b-√Δ)/2a=(3-√23)/7
x(2)=(-b+√Δ)/2a=(3+√23)/7
On en déduis alors: S={(3-√23)/7;(3+√23)/7}