Question

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

Answer 1

(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)

Answer 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 

 

 

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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