Bonjour, aidez moi avec cet exercice svp j’ai de grande difficulté, merci infiniment !

Réponse :

a) (5x-3)(x-5)= (2x+5)²+90

5x²-25x-3x+15 = 4x²+20x+25+90

5x²-28x+15=4x²+20x+115

5x²-28x+15-4x²-20x-115=0

x²-48x-100=0

b²-4ac = (-48)²-4(1*-100)=2704

Δ>0  2solutions

(-b-√Δ)/2a=(48-52)/2 =-2

(-b+√Δ)/2a = (48+52)/2=50

S{-2;50}

x+[tex]\frac{1}{x}[/tex]=2

x≠0

x+ 1/x - 2 =0

(x²+1-2x)/x=0

x²-2x+1=0

(x-1)²=0

x=1

3x+5/2x+1 = 4x+7/x-1

x≠-1/2;1

(3x+5)(x-1) = (4x+7)(2x+1)

(3x+5)(x-1) -(4x+7)(2x+1)=0

3x²-3x+5x-5-(8x²+4x+14x+7)=0

3x²+2x-5-8x²-18x-7=0

-5x²-16x-12=0

b²-4ac= (-16)²-4(-5*-12) =16

Δ>0  2solutions

(-b-√Δ)/2a =(16-4)/-10 =12/-10= -6/5

(-b+√Δ)/2a=(16+4)/-10 = -2

S{-6/5;-2}

3x²-27/2x²+6x=0

x≠-3;0

3(9x²-9)/2x(x+3)=0

3(3x-3)(3x+3)/2x(x+3)=0

3(x-3)/2x=0

x=3

x-5/x-1 + x-3/x-2 = -9x+17/(x-1)(x-2)

x≠1;2

x-5/x-1 + x-3/x-2 -(-9x+17/(x-1)(x-2))=0

[(x-5)(x-2)+(x-3)(x-1)+9x-17]/(x-1)(x-2)=0

(x²-5x-2x+10+x²-3x-x+3+9x-17)/(x-1)(x-2)=0

(2x²-2x-4)/(x-1)(x-2)=0

2x²-2x-4=0

b²-4ac=(-2)²-4(2*-4)=36

Δ>0  2solutions

(-b+√Δ)/2a =(2+6)/4 =2

(-b-√Δ)/2a=(2-6)/4 =-1

S{2;-1}

Explications étape par étape :