Matematyka

Autorzy:Marcin Kurczab Elżbieta Kurczab Elżbieta Świda

Wydawnictwo:Krzysztof Pazdro

Rok wydania:2014

Rozwiąż nierówności 4.38 gwiazdek na podstawie 8 opinii
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`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)=`

`\ \ \ \ \ \ 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}))`