a) 4x(x + 1) = 8(x + 1) g) 5x(x – 2000) – x + 2000 = 0 b) x(x – 1) – 2(1 – x) = 0 h

a) 4x(x + 1) = 8(x + 1) g) 5x(x – 2000) – x + 2000 = 0 b) x(x – 1) – 2(1 – x) = 0 h) x2 – 4x = 0 c) 2x(x – 2) – (2 – x)2 = 0 k) (1 – x)2 – 1 + x = 0 d) (x – 3)3 + 3 – x = 0 m) x + 6×2 = 0 e) 5x(x – 2) – (2 – x) = 0 n) (x + 1) = (x + 1)2
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  1. Đáp án:

    a) 4x(x + 1) = 8(x + 1)

    4x(x + 1)+8(x + 1)=0

    (x + 1).(4x+8)=0

    TH1:x+1=0

           x    =0-1

           x=-1

    TH2:4x+8=0

           4x     =0-8

           4x     =-8

            x      =-8:4

            x=-2

    Vậy x∈{-1;-2}

    b) x(x – 1) – 2(1 – x) = 0

    x(x – 1) + 2(x – 1) = 0

    (x – 1).(x+2)=0

    TH1:x – 1=0

           x      =0+1

          x        =1

    TH2:x+2=0

           x     =0-2

          x      =-2

    Vậy x ∈{1;-2}

    c) 2x(x – 2) – (2 – x)2 = 0

    2x(x – 2) +(x – 2)2 =0

    (x – 2).(2x+2)=0

    TH1:x-2=0

          x     =0+2

          x     =2

    TH2:2x+2=0

           2x     =0-2

           2x     =-2

             x     =-2:2

            x      =-1

    Vậy x∈{2;-1}

    d) (x – 3)3 + 3 – x = 0

    (x – 3)3 – (x-3)=0

    (x – 3).(3-1)=0

    TH1: x-3=0

            x    =0+3

           x     =3

    TH2:3-1=0(vô lí)

    Vậy x =3

    e) 5x(x – 2) – (2 – x) = 0

    (x – 2).(5x-1)=0

    TH1:x-2=0

           x    =0+2

          x     =2

    TH2:5x-1=0

           5x    =0+1

           5x     =1

             x     =1:5

             x      =1/5

    Vậy x∈{2;1/5}

    g) 5x(x – 2000) – x + 2000 = 0

    5x(x – 2000) –( x – 2000)=0

    (x – 2000).(5x-1)=0

    TH1: x-2000=0

            x          =0+2000

            x           =2000

    TH2:5x-1=0

           5x    =0+1

           5x     =1

             x     =1:5

             x      =1/5

    Vậy x∈{2000;1/5}

    h) x2 – 4x = 0

    x.(2-4)=0

    x.(-2)=0(vô lí)

    vậy x ∈∅

    k) (1 – x)2 – 1 + x = 0

    (1 – x)2 –( 1 – x)=0

    (1 – x).(2-1)=0

    TH1:1-x=0

              x =1-0

              x  =1

    TH2:2-1=0(vô lí)

    vậy x =1

    m) x + 6×2 = 0

         x+  12x  =0

         x.(1+12) =0

         x.13       =0(vô lí)

    vậy x ∈∅

    n) (x + 1) = (x + 1)2

    (x + 1) +(x + 1)2=0

    (x + 1).(1+2)=0

    (x + 1).3=0

    x=-1

    :xong rồi đó bạn

     

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