TRẢ LỜI

1. d)

   (x-2)^2-(x-3).(x+3)=6

   =>x^2-2x.2+2^2-(x^2-3^2)=6

   =>x^2-4x+4-x^2+9=6

   =>(x^2-x^2)-4x+4=-3

   =>-4x=-7

   =>x=\frac{7}{4}

   Vậy x=\frac{7}{4}

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   e)

   4x.(x-3)-2x+6=0

   =>4x.(x-3)-(2x-6)=0

   =>4x.(x-3)-2.(x-3)=0

   =>(x-3).(4x-2)=0

   =>\(\left[ \begin{array}{l}x-3=0\\4x-2=0\end{array} \right.\) 

   =>\(\left[ \begin{array}{l}x=3\\4x=2\end{array} \right.\) 

   =>\(\left[ \begin{array}{l}x=3\\x=\dfrac{2}{4}\end{array} \right.\) 

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

   Vậy x=3; x=\frac{1}{2}

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   f)

   x.(2x-3)-2.(3-2x)=0

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

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

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

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

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

   Vậy x=\frac{3}{2}; x=-2

    

   Trả lời
2. d) ( x –  2 )^2 – ( x – 3 ) ( x + 3 ) =6

   ⇒ x^2 – 4x + 4 – x^2 + 9 =6

   ⇒ -4x + 13 =6

   ⇒ -4x = -7

   ⇒ x = 7/4

   Vậy x = 7/4

   e) 4x ( x – 3 ) -2x + 6 =0

   ⇒ 4x ( x – 3 ) – 2 ( x – 3 )=0

   ⇒ ( x – 3 ) ( 4x – 2 ) =0

   ⇒ \(\left[ \begin{array}{l}x – 3=0\\4x – 2=0\end{array} \right.\) 

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

   Vậy x ∈ { 3 ; \frac{1}{2} }

   f) x ( 2x – 3) – 2 ( 3 -2x ) =0

   ⇒ x ( 2x – 3 ) + 2 ( 2x – 3 ) =0

   ⇒ ( x + 2 ) ( 2x – 3 ) =0 

   ⇒ \(\left[ \begin{array}{l}x + 2=0\\2x – 3=0\end{array} \right.\) 

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

   Vậy x ∈ { -2 ; 3/2 }

   Trả lời