Toán Lớp 8: B=( ($\frac{x+1}{1-x}$- $\frac{1-x}{1+x}$ $\frac{4x^{2}}{x{2}-1}$): $\frac{4x^{2}-4}{x^{2}-2x+1}$ Rút gọn
Rút gọn:B= ($\dfrac{x+1}{1-x}$- $\dfrac{1-x}{1+x}$- $\dfrac{4x²}{x²-1}$) : $\dfrac{4x²-4}{x²-2x+1}$ = ($\dfrac{-x-1}{x-1}$- $\dfrac{1-x}{x+1}$- $\dfrac{4x²}{(x-1)(x+1)}$) : $\dfrac{4(x²-1)}{(x-1)²}$ = $\dfrac{(-x-1)(x+1)-(1-x)(x-1)-4x²}{(x-1)(x+1)}$ : $\dfrac{4(x-1)(x+1)}{(x-1)²}$ = $\dfrac{-x²-x-x-1-x+1+x²-x-4x²}{(x-1)(x+1)}$ : $\dfrac{4(x-1)(x+1)}{(x-1)²}$ = $\dfrac{-4x²-4x}{(x-1)(x+1)}$ : $\dfrac{4(x-1)(x+1)}{(x-1)²}$ = $\dfrac{-4x(x+1)}{(x-1)(x+1)}$ . $\dfrac{(x-1)²}{4(x-1)(x+1)}$ = $\dfrac{x}{x+1}$ Chúc bạn học tốt^^ Trả lời
B= ($\dfrac{x+1}{1-x}$- $\dfrac{1-x}{1+x}$- $\dfrac{4x²}{x²-1}$) : $\dfrac{4x²-4}{x²-2x+1}$
= ($\dfrac{-x-1}{x-1}$- $\dfrac{1-x}{x+1}$- $\dfrac{4x²}{(x-1)(x+1)}$) : $\dfrac{4(x²-1)}{(x-1)²}$
= $\dfrac{(-x-1)(x+1)-(1-x)(x-1)-4x²}{(x-1)(x+1)}$ : $\dfrac{4(x-1)(x+1)}{(x-1)²}$
= $\dfrac{-x²-x-x-1-x+1+x²-x-4x²}{(x-1)(x+1)}$ : $\dfrac{4(x-1)(x+1)}{(x-1)²}$
= $\dfrac{-4x²-4x}{(x-1)(x+1)}$ : $\dfrac{4(x-1)(x+1)}{(x-1)²}$
= $\dfrac{-4x(x+1)}{(x-1)(x+1)}$ . $\dfrac{(x-1)²}{4(x-1)(x+1)}$
= $\dfrac{x}{x+1}$
Chúc bạn học tốt^^