tìm x biết: 1) a) (x+1)(x+3)-x(x-1)=8 b) 9x^2= 1-(3x+1)(2x-9)

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1. a)(x+1)(x+3)-x(x-1)=8

   x² +4x+3- x²+x=8

   5x=5

   x=1

   Vậy x=1

   b) 9x²= 1-(3x+1)(2x-9)

   9x²=1- 6x² +27x -2x +9

   15x²-25x-10=0

   x²-$\frac{5}{3}$x -$\frac{2}{3}$ =0

   x² -2x +$\frac{1}{3}$x -$\frac{2}{3}$=0

   x(x-2)+$\frac{1}{3}$(x-2)=0

   (x+$\frac{1}{3}$).(x-2)=0

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

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

   Vậy \(\left[ \begin{array}{l}x=-\frac{1}{3}\\x=2\end{array} \right.\)

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2. Đáp án:

   Bài 1: giải các pt sau:

   a,(x-1)^2-(x+1)^2=2(x+3)

   (2x-1)^2-(2x+1)^2=4(x-3)

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

   b,9x^12-(x+1)^3=3x-3

   (3x-2).(9x^2+6x+4)-(3x+1).(9x^2-3x+1)=(2x+1).(2x-1)-4x(x-3)

   Giải thích các bước giải:

    

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