Bài 1.Tìm x khác Z a)3x-4/x+1 khác Z b)x²-7/x-2 khác Z c)4x-7/2x+1 khác Z d)x²+3x-2/x+2 khác Z

22/05/2023

1 bình luận về “Bài 1.Tìm x khác Z a)3x-4/x+1 khác Z b)x²-7/x-2 khác Z c)4x-7/2x+1 khác Z d)x²+3x-2/x+2 khác Z </p”

1. a,

   Ta có: (3x – 4)/(x + 1) = (3x + 3 – 7)/(x + 1) = (3(x + 1) – 7)/(x + 1) =  3 – 7/(x + 1)

   Để (3x – 4)/(x + 1) \ne Z thì

   7 \cancel{\vdots} x + 1

   => x + 1 \notin Ư(7)

   => x + 1 \notin {+-1,+-7}

   => x \notin {-8,-2,0,6}

   Vậy x \notin {-8,-2,0,6} thì (3x- 4)/(x + 1) \ne Z

   b,

   Ta có: (x^2 – 7)/(x – 2) = (x^2 – 4 – 3)/(x – 2) = ((x – 2)(x + 2) – 3)/(x – 2) = x + 2 – 3/(x – 2)

   Để (x^2 – 7)/(x – 2) \ne Z thì

   3 \cancel{\vdots} x – 2

   => x – 2 \notin Ư(3)

   => x – 2 \notin {+-1,+-3}

   => x  \notin {-1,1,3,5}

   Vậy x \notin {-1,1,3,5} thì (x^2 – 7)/(x – 2) \ne Z

   c,

   Ta có: (4x – 7)/(2x + 1) = (4x + 2- 9)/(2x + 1) = (2(2x + 1) – 9)/(2x + 1) = 2 – 9/(2x + 1)

   Để (4x – 7)/(2x + 1) \ne Z thì

   9 \cancel{\vdots} 2x + 1

   => 2x + 1 \notin Ư(9)

   => 2x + 1 \notin {+-1,+-3,+-9}

   => 2x \notin {-10,-4,-2,0,2,8}

   => x \notin {-5,-2,-1,0,1,4}

   Vậy x \notin {-5,-2,-1,0,1,4} thì (4x – 7)/(2x + 1) \ne Z

   d,

   Ta có: (x^2 + 3x – 2)/(x + 2) = (x^2 + 2x + x + 2 – 4)/(x + 2) = (x(x + 2) + (x + 2) – 4)/(x + 2) = ((x + 2)(x + 1) – 4)/(x + 2) = x + 1 – 4/(x + 2)

   Để (x^2 + 3x – 2)/(x + 2) \ne Z thì

   4 \cancel{\vdots} x + 2

   => x + 2 \notin Ư(4)

   => x + 2 \notin {+-1,+-2,+-4}

   =>  x \notin {-6,-4,-3,-1,0,2}

   Vậy x \notin {-6,-4,-3,-1,0,2} thì (x^2 +3x + 2)/(x + 2) \ne Z

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