Bardzo proszę o rozwiązanie zad zadania jak najszybciej !

A) (3x^2+4x-4) / x^2+x-4=>1 //*x^2+x-2

3x^2+4x-4>x^2+x-2
2x^2+3x-2>0
D=b^2-4ac=9+16=25
\D=5

x1=(-b-\D)/2a=(-3-5)/2*2=-8/4=-2
x2=(-b-\D)2a=(-3+\D)2a=(-3+5)/2*2=2/2=1


B)4/(x+3) + 2/(x-3) <(5x-1)(x^2-9) //*(x^2-9)

4(x-3)+2(x+3)<5x-1
4x-12+2x+6<5x-1
x-5<0
x<5

A)(x^2+1)/(x-1)=2/(x-1) //*(x-1)
x^2+1=2
x^2=2-1
x^2=1
x1=1 ^ x2=-1

B)(x+1)/(x-1)+1=2x //*(x-1)

(x+1)+(x-1)=2x(x-1)
x+1+x-1=2x^2-2x
-2x^2+4x=0 //(-2)
x^2-2x=0

x(x-2)=0 ,x1=0
x-2=0 x2=2