Toán Lớp 8: Rút gọn: `(x^2 + 2xy + y – 1)/(x^2 + 2x + 1 – y^2)`

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1. $\text{ @HV }$

   $\dfrac{x^2 + 2xy + y^2 – 1 }{x^2 + 2x + 1 – y^2}$

   = $\dfrac{(x + y)^2 – 1^2}{(x + 1)^2 – y^2}$

   = $\dfrac{(x + y – 1).(x + y + 1)}{(x + 1 – y).(x + 1 + y)}$

   = $\dfrac{x + y – 1}{x + 1 – y}$

   = $\dfrac{x + y – 1}{x – y + 1}$

    

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2. $\dfrac{x^2+2xy+y^2-1}{x^2+2x+1-y^2}$

   $=\dfrac{(x+y)^2-1^2}{(x+1)-y^2}$

   $=\dfrac{(x+y-1)(x+y+1)}{(x+1-y)(x+1+y)}$

   Có $(x+y+1)$ chung

   $=\dfrac{x+y-1}{x+1-y}$

   $#thanhmaii2008$

    

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