2° - Matemática

10/02/15

1) Resolva as seguintes situações:

    a) (2+3)²=(2+3)x(2+3)=25

    b) (2+3)³=(2+3)x(2+3)x(2+3)=125

    c) (2+a)²=(2+a)x(2+a)=4+2a+2a+a²=4+4a+a²

2) Resolva:

    a) x²+2x+1=0

        ∆=b²-4ac

        ∆=2²-4.1.1

        ∆=4-4=0

                                -> -2=-1

                                     2

        x=-2±V0=-2±0

               2.1       2

                              -> -2=-1

                                   2

       V={-1}

    b) f(x)=x²-5x+6

        a>0 U

        x²-5x+6=0

        ∆=b²-4ac

        ∆=-5²-4.1.6

        ∆=25-24=1

                                -> 6=3

                                     2

        x=5±V1=5±1

              2.1      2

                              -> 4=2

                                   2

         S={2,3}

        V=(-b, -∆)

            ( 2a 4a)

       (0,y)

       x²+5x+6=0

       0²-5(0)+6=6