ex 1: a) x+4√x -5 =0 ; donc: x-5 =-4√x
(x-5)² = (-4√x)²
x²-10x+25 =16x
x²-10x+25-16x=0
x²-26x+25 =0
Δ=(-26)² - 4(1)(25)
Δ=676-100=576
X1=(-(-26)+√576)/2 = (26+24)/2 = 50/2 =25
X2=(26-24)/2 = 2/2 =1 ;... solutions : S ={1 ; 25}
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b) (x²-2x)/(x²+4) > 1 ( x²+4 strictement positif, )
donc: x²-2x > x²+4
x²-x² -2x > 4
-2x> 4
x<4/-2
x<-2 ;;; solutions : S= ]-∞ ; -2[
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c) 5/(x-6) < 3x-2
si : x >6 ; 5 < (x-6)(3x-2)
(x-6)(3x-2) > 5
3x²-2x-18x+12>5
3x²-20x+12-5>0
3x²-20x+7>
Δ=(-20)²-4(7)(3) =400-84=316
X1=(20+√316)/(2×3) =(20+2√79)/6 =(10+√79)/3 ; X2 = (10-√79)/3
tableau de signes ......................................................................................
x -∞ (10-√79)/3 (10+√79)/3 +∞
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3x²-20x+7 + 0 - 0 +
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x>6 et 3x²-20x+7>0 ⇔ x∈] -∞ ; (10-√79)/3[∪](10+√79)/3 ; +∞[
et x∈]6 ; +∞[
donc x∈ ](6+√79)/3 ; +∞[
x<6 et 3x²-20x+7< 0 ⇔ x∈ ](10-√79)/3 ; (10+√79)/3[ et x∈]-∞ ; 6[
donc :x∈ ](10-√79)/3 ; 6[
solutions : S=](10-√79)/3 ; 6[∪] (6+√79)/3 ; + ∞[
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d) -6/(x²-9) +2=1/(x-3)
-6/(x²-9) + 2(x²-9)/(x²-9) = 1(x+3)/(x²-9)
-6+2(x²-9) =x+3
-6+2x²-18=x+3
2x²-24-x-3=0
2x²-x-27=0
Δ= (-1)² -4(2)(-27)
Δ= 1+216=217
X1 =(1+√217)/4 ; X2 = (1-√217)/4
solution : S= { (1+√217)/4 ; (1-√217)/4}
