Matematyka

Autorzy:Marcin Kurczab Elżbieta Kurczab Elżbieta Świda

Wydawnictwo:Krzysztof Pazdro

Rok wydania:2014

Rozwiąż nierówności 4.38 gwiazdek na podstawie 8 opinii
  1. Liceum
  2. 2 Klasa
  3. Matematyka

`a)` 

`|x^2-4x-3|<6` 

`x^2-4x-3<6\ \ \ |-6 \ \ \ \ wedge\ \ \ \ x^2-4x-3> -6\ \ \ |+6` 

`^"(1)" x^2-4x-9<0 \ \ \ \ \ \ \ \ \ wedge\ \ \ \ ^"(2)" x^2-4x+3>0` 

 

`(1)\ \ \ x^2-4x-9<0` 

`\ \ \ \ \ \ Delta=(-4)^2-4*1*(-9)=` `16+36=52` 

`\ \ \ \ \ \ sqrtDelta=sqrt52=sqrt4*sqrt13=2sqrt13` 

`\ \ \ \ \ \ x_1=(4-2sqrt13)/2=2-sqrt13` 

`\ \ \ \ \ \ x_2=2+sqrt13` 

 

           

`\ \ \ \ \ \ x in (2-sqrt13;\ 2+sqrt13)` 

`\ \ \ \ \ \ 2-sqrt13~~2-3,6=-1,6`  

`\ \ \ \ \ \ 2+sqrt13~~2+3,6=5,6` 

 

 

 

`(2)\ \ \ x^2-4x+3>0` 

`\ \ \ \ \ \ (x-3)(x-1)>0` 

            

` \ \ \ \ \ \ x in (-infty,\ 1)\ uu\ (3,\ +infty)` 

 

 

 

`\ \ \ \ \ \ ((1)\ \ \ wedge\ \ \ (2))\ \ \ =>\ \ \ (x in (2-sqrt13,\ 2+sqrt13)\ \ \ \ wedge\ \ \ x in (-infty,\ 1)\ uu\ (3,\ +infty))\ \ \ =>\ \ \ ul(ul(x in (2-sqrt13,\ 1)\ uu\ (3,\ 2+sqrt13)))`  

 

 

 

 

 

 

`b)` 

`x+1=0\ \ \ <=>\ \ \ x=-1`  

`x-1=0\ \ \ <=>\ \ \ x=1` 

 

`"Rozpatrujemy trzy przypadki:"`  

`^"(1)"\ x in (-infty,\ -1)\ \ \ vee\ \ \ ^"(2)"\ x in <<-1,\ 1)\ \ \ vee\ \ \ ^"(3)"\ x in <<1,\ +infty)` 

 

 

`(1)\ x in (-infty,\ -1)` 

`\ \ \ \ \ -(x+1)>=x^2+(x-1)` 

`\ \ \ \ \ -x-1>=x^2+x-1\ \ \ |+x+1` 

`\ \ \ \ \ x^2+2x<=0` 

`\ \ \ \ \ x(x+2)<=0` 

    

        

`\ \ \ \ (x in <<-2,\ 0>>\ \ \ wedge\ \ \ x in (-infty,\ -1))\ \ \ =>\ \ \ x in <<-2,\ -1)` 

 

 

`(2)\ x in <<-1,\ 1)` 

`\ \ \ \ \ x+1>=x^2+x-1\ \ \ |-x-1` 

`\ \ \ \ \ x^2-2<=0` 

`\ \ \ (x-sqrt2)(x+sqrt2)<=0` 

 

         

`\ \ \ \ \ (x in <<-sqrt2,\ sqrt2>>\ \ \ wedge\ \ \ x in <<-1,\ 1))\ \ \ =>\ \ \ x in <<-1,\ 1)`  

 

 

`(3)\ x in<<1,\ +infty)` 

`\ \ \ \ \ x+1>=x^2-x+1\ \ \ |-x-1`  

`\ \ \ \ \ x^2-2x<=0` 

`\ \ \ \ \ x(x-2)<=0` 

         

`\ \ \ \ \ (x in <<0,\ 2>>\ \ \ wedge\ \ \ x in <<1,\ +infty))\ \ \ =>\ \ \ x in <<1,\ 2>>` 

 

 

`\ \ \ \ \ ((1)\ \ \ vee\ \ \ (2)\ \ \ vee\ \ \ (3))\ \ \ =>\ \ \ (x in <<-2,\ -1)\ \ \ vee\ \ \ x in <<-1,\ 1)\ \ \ vee\ \ \ x in <<1,\ 2>>)\ \ \ =>\ \ \ ul(ul(x in <<-2,\ 2>>))` 

 

 

 

 

 

`c)` 

`1-x^2=0\ \ \ <=>\ \ \ x=1\ \ \ vee\ \ \ x=-1` 

 

`"Rozpatrujemy cztery przypadki:"` 

`^"(1)" \ x in (-infty,\ -1)\ \ \ vee\ \ \ ^"(2)"\ x in <<-1,\ 0)\ \ \ vee\ \ \ ^"(3)"\ x in <<0,\ 1)\ \ \ vee\ \ \ ^"(4)"\ x in\ <<1,\ +infty)` 

 

 

`(1)\ x in (-infty,\ -1)` 

`\ \ \ \ \ -x-1+x^2>1\ \ \ |-1` 

`\ \ \ \ \ x^2-x-2>0`   

`\ \ \ \ \ (x-2)(x+1)>0` 

      

` \ \ \ \ \ (x in (-infty,\ -1)\ uu\ (2,\ +infty)\ \ \ wedge\ \ \ x in (-infty,\ -1))\ \ \ =>\ \ \ x in (-infty,\ -1)` 

 

 

`(2)\ x in <<-1,\ 0)` 

`\ \ \ \ \ -x+1-x^2>1\ \ \ |-1` 

`\ \ \ \ \ -x^2-x>0\ \ \ |*(-1)` 

`\ \ \ \ \ x^2+x<0` 

`\ \ \ \ \ x(x+1)<0` 

 

       

`\ \ \ \ \ (x in (-1,\ 0)\ \ \ wedge\ \ \ x in <<-1,\ 0))\ \ \ =>\ \ \ x in (-1,\ 0)` 

 

 

`(3)\ x in <<0,\ 1)` 

`\ \ \ \ \ x+1-x^2>1\ \ \ |-1` 

`\ \ \ -x^2+x>0\ \ \ |*(-1)` 

`\ \ \ x^2-x<0`  

`\ \ \ x(x-1)<0`   

     

`\ \ \ \ \ (x in (0,\ 1)\ \ \ wedge\ \ \ x in <<0,\ 1))\ \ \ =>\ \ \ x in (0,\ 1)` 

 

 

`(4)\ x in <<1,\ +infty)` 

`\ \ \ \ \ \ x-1+x^2>1\ \ \ |-1` 

`\ \ \ \ \ x^2+x-2>0` 

`\ \ \ \ \ (x+2)(x-1)>0` 

         

`\ \ \ \ \ (x in (-infty,\ -2)\ uu\ (1,\ +infty)\ \ \ wedge\ \ \ x in <<1,\ +infty))\ \ \ =>\ \ \ x in (1,\ +infty)` 

 

 

`\ \ \ \ \ ((1)\ \ \ vee\ \ \ (2)\ \ \ vee \ \ \ (3)\ \ \ vee\ \ \ (4))\ \ \ =>\ \ \ (x in (-infty,\ -1)\ \ \ vee\ \ \ x in (-1,\ 0)\ \ \ vee\ \ \ x in (0,\ 1)\ \ \ vee\ \ \ x in (1, +infty))\ \ \ =>\ \ \ ul(ul(x in RR-{-1,\ 0,\ 1}))`