Giải câu 1 trang 60 toán VNEN 8 tập 1.
a) ($\frac{2x + 1}{2x – 1}$ - $\frac{2x – 1}{2x + 1}$) : $\frac{4x}{10x – 5}$ = $\frac{(2x + 1)^{2} – (2x – 1)^{2}}{(2x – 1)(2x + 1)}$ : $\frac{4x}{5(2x – 1)}$ = $\frac{8x}{(2x – 1)(2x + 1)}$ . $\frac{5(2x – 1)}{4x}$ = $\frac{10}{2x + 1}$;
b) ($\frac{1}{x^{2} + 1}$ - $\frac{2 – x}{x + 1}$) : ($\frac{1}{x}$ + x – 2) = $\frac{x + 1 – (2x^{2} - x^{3} + 2 - x}{(x^{2} + 1)(x + 1)}$ : $\frac{1 + x^{2} – 2x}{x}$ = $\frac{x^{3} – 2x^{2} + 2x - 1}{(x^{2} + 1)(x + 1)}$ . $\frac{x}{(x – 1)^{2}}$
= $\frac{(x – 1)^{3}}{(x^{2} + 1)(x + 1)}$ . $\frac{x}{(x – 1)^{2}}$ = $\frac{x(x – 1)}{(x^{2} + 1)(x + 1)}$;
c) $\frac{1}{x – 1}$ - $\frac{x^{3} - x}{x^{2} + 1}$ . ($\frac{1}{x^{2} – 2x + 1}$ - $\frac{1}{1 - x^{2}}$) = $\frac{1}{x - 1}$ - $\frac{x^{3} - x}{x^{2} + 1}$.$\frac{1 - x^{2} + x^{2} – 2x + 1}{(x – 1)^{2}.(1 - x^{2})}$ = $\frac{1}{x - 1}$ - $\frac{2(x – 1)}{(x – 1)^{2}.(x^{2} – 1)}$.$\frac{x^{3} - x}{x^{2} + 1}$
= $\frac{1}{x – 1}$ - $\frac{2(x – 1)}{(x – 1)^{2}.(x^{2} – 1)}$ . $\frac{x(x^{2} – 1)}{x^{2} + 1}$ = $\frac{1}{x - 1}$ - $\frac{2x}{(x – 1)(x^{2} + 1)}$ = $\frac{x^{2} + 1 – 2x}{(x – 1)(x^{2} + 1)}$ = $\frac{x – 1}{x^{2} + 1}$.
d) ($\frac{x^{2} + xy}{x^{3} + x^{2}y + xy^{2} + y^{3}}$ + $\frac{y}{x^{2} + y ^{2}}$) : ($\frac{1}{x – y}$ - $\frac{2xy}{x^{3} - x^{2}y + xy^{2} – y^{3}}$) = ($\frac{x(x + y)}{(x^{2} + y^{2})(x + y)}$ + $\frac{y}{x^{2} + y^{2}}$) : ($\frac{1}{x - y}$ - $\frac{2xy}{(x – y)(x^{2} + y^{2})}$)
= $\frac{x + y}{x^{2} + y^{2}}$ : $\frac{(x – y)^{2}}{(x – y)(x^{2} + y^{2})}$ = $\frac{x + y}{x^{2} + y^{2}}$ . $\frac{x^{2} + y^{2}}{x - y}$ = $\frac{x + y}{x - y}$.