Решите плз (2x^2+3x)/(3-x)=(x-x^2)/(x-3) (5x-7)/(x-3)=(4x-3)/x

(2x²+3x)/(3-x)=(x-x²)/(x-3), x≠3 (2x²+3x)/(3-x)-(x-x²)/(x-3)=0 (2x²+3x)/(3-x)+(x-x²)/(3-x)=0 |*(3-x) 2x²+3x+x-x²=0 x²+4x=0 x(x+4)=0 x=0 x=-4 Ответ: 0; -4 (5x-7)/(x-3)=(4x-3)/x |*x(x-3) x(5x-7)-(x-3)(4x-3)=0 5x²-7x-(4x²-3x-12x+9)=0 5x²-7x-4x²+3x+12x-9=0 x²+8x-9=0 a=1, b=8, c=-9 D/4 = (b/2)²-ac D/4 = 16+9 = 25, √D/4 = 5 x= (-b/2+-√D/4)/a x₁ = -4+5 = 1 x₂ = -4-5 = -9 Ответ: 1;-9

1 (2x²+3x)/(3-x)=(x²-x)/(3-x) x≠3 2x²+3x=x²-x x²+4x=0 x(x+4)=0 x1=0 x2=-4 2 (5x-7)/(x-3)-(4x-3)/x=0 x≠0 x≠3 (5x-7)x-(4x-3)(x-3)=0 5x²-7x-4x²+12x+3x-9=0 x²+8x-9=0 x1+x2=-8 U x1*x2=-9 x1=-9 x2=1