(x-5)(x+2)+(x-5)(2x+1)=0 (x+3)²-(2x-3)²=0 x²+2x+1=(x+1)(2x-3)

(x-5)(x-2)+(x-5)(2x+1)=(x-5)(x+2+2x+1)=(x-5)(3x+3)=3(x-5)(x+1)
Donc (x-5)(x+1)=0 ssi x=5 ou x=-1

(x+3)²-(2x-3)²=(x+3+2x-3)(x+3-2x+3)=3x(6-x)
Donc 3x(6-x)=0 ssi x=0 ou x=6

x²+2x+1=(x+1)(2x-3) <=> (x+1)²-(x+1)(2x-3)=0
<=> (x+1)(x+1-2x+3)=0
<=> (x+1)(4-x)=0
<=> x=-1 ou x=4