zadanie w załaczniku zad 2 i 3 . bardzo blagam o rozwiazanie :( dam naj i bede bardzo wdzieczna;(

(x-3)(x+3) = 3x - 5 x^2 - 9 = 3x - 5 x^2 - 3x -4 = 0 delta = 9 +16 = 25 x1 = (3-5)/2 = -1 x2 = (3+5)/2 = 4   2x + 1 = 12x + 24 10x = -23 x = -2.3   (3^(4/3) * 3^(-2) * 3^4 ) / ( 3^(4/3) * 3^(1/2) )= 3^(3/2) 3^(3/2) = 3^(3/2)

(x-3):(3x-5)=1:(x+3) (x-3)(x+3)=3x-5 x² -9-3x+5=0   x²-3x-4=0    Δ =9+16=25 √Δ =5    x1=(3-5):2=-2:2=-1 x2=(3+5):2=8:2=4  -------------------------------------    (2x+1):(x+2)=12  2x+1=12(x+2) 2x+1=12x+24 12x-2x=1-24 10x=-23 x=-2,3      ------------------------------------------ L=(9^2/3*3^-2*27^4/3):(81^1/3*√3)=[(3^2)^2/3*3^-2*(3^3)^4/3]:[(3^4)^1/3*3^1/2]  =[3^4/3*3^-2*3^4]:[3^4/3*3^1/2]=[3^-2*3^4]:[ 3^1/2]=[3^-2+4]:3^1/2=3^2:3^1/2=3^2-1/2=3^3/2     P=9^3/4=(3^2)^3/4=3^6/4=3^3/2   L=P   ^-oznacza do potęgi /-kreska ułamkowa