0 bình luận về “Môn Toán Lớp 8: a,2x-3=0 b,3x+2=8-x-7x c,x²-9=(2x+1)(x-3) d,4x/3+5x/2=1 e,3/x+4x+1/x-2=12x-10/x(x-2)”

1. a) 2x – 3 = 0

   <=> 2x = 3

   <=> x = 3/2

   Vậy S = { 3/2 }

   b) 3x + 2 = 8 – x – 7x

   <=> 3x + 2 = 8 – 9x

   <=> 3x + 9x = 8 – 2

   <=> 12x = 6

   <=> x = 1/2
   Vậy S = { 1/2 }

    c) x^2 – 9 = ( 2x + 1 ) ( x – 3 )

   <=> (  x- 3 ) ( x + 3 ) – ( 2x + 1 ) ( x – 3 ) = 0

   <=> ( x – 3 ) ( x + 3 – 2x – 1 ) = 0

   <=> ( x – 3 ) ( x – 2 ) = 0

   <=>\(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\) 

   <=>\(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\) 

   Vậy S = { 3 ; 2 }

   d) ( 4x )/3 + ( 5x )/2 = 1

   => 8x + 15x = 6

   <=> 23x = 6

   <=> x = 6/23

   Vậy S= { 6/23 }

   e) 3/(x + 4x ) + 1/( x – 2 ) = ( 12x – 10 )/( x ( x – 2 ))

   => 3 ( x – 2 ) + x ( x + 2 ) = ( 12x – 10 ) ( x + 2 )

   <=> 3x – 6 + x^2 + 2x = 12x^2 + 24x – 10x – 20

   <=> x^2 + 5x – 6 = 12x^2 + 14x – 20

   <=> x^2 – 12x^2 + 5x – 14x = – 20 + 6

   <=> – 11x^2 – 9x = –  14

   <=> – 11x^2 – 9x + 14 = 0

   <=> 11x^2 + 9x – 14 = 0

   Vậy S= { ∅ }

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