giải phương trình a) ( 3x – 2 ) ( 2x + 1 ) = ( 2x + 1 )^2 b) 2 / x – 2 – 3 / 3 – x = 3x – 20 / ( x – 3 ) ( x – 2 )

1 bình luận về “giải phương trình a) ( 3x – 2 ) ( 2x + 1 ) = ( 2x + 1 )^2 b) 2 / x – 2 – 3 / 3 – x = 3x – 20 / ( x – 3 ) ( x – 2 )”

1. \bb a)

   (3x-2).(2x+1)=(2x+1)^2

   <=>6x^2+3x-4x-2=(2x)^2+2.2x.1+1^2

   <=>6x^2-x-2=4x^2+4x+1

   <=>6x^2-4x^2-x-4x-2-1=0

   <=>2x^2-5x-3=0

   <=>2x^2+x-6x-3=0

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

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

   <=>\(\left[ \begin{array}{l}2x+1=0\\x-3=0\end{array} \right.\) 

   <=>\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=3\end{array} \right.\) 

   Vậy S={-1/2; 3}

   $\\$

   \bb b)

   2/(x-2)-3/(3-x)=(3x-20)/((x-3).(x-2))(ĐK:x\ne3; x\ne2)

   <=>(2.(x-3)+3.(x-2))/((x-2).(x-3))=(3x-20)/((x-3).(x-2))

   =>2.(x-3)+3.(x-2)=3x-20

   <=>2x-6+3x-6=3x-20

   <=>5x-12-3x+20=0

   <=>2x+8=0

   <=>2x=-8

   <=>x=-4(tm)

   Vậy S={-4}

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