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Đáp án + Giải thích các bước giải:
Bài 1:
a) $A = \dfrac{x - 1}{x} - \dfrac{x + 1}{x^2 - x} + \dfrac{3(x - 1)}{x^2 - 2x + 1}(x \ne 0; \pm 1)$
$= \dfrac{x - 1}{x} - \dfrac{x + 1}{x(x - 1)} + \dfrac{3(x - 1)}{(x - 1)^2}$
$= \dfrac{x - 1}{x} - \dfrac{x + 1}{x(x - 1)} + \dfrac{3}{x - 1}$
$= \dfrac{(x - 1)^2}{x(x - 1)} - \dfrac{x + 1}{x(x - 1)} + \dfrac{3x}{x(x - 1)}$
$= \dfrac{x^2 -2x + 1 - x - 1 + 3x}{x(x - 1)}$
$= \dfrac{x^2}{x(x - 1)}$
$= \dfrac{x}{x - 1}$
b) $B = \dfrac{x + y}{2(x - y)} - \dfrac{x - y}{2(x + y)} + \dfrac{2y^2}{x^2 - y^2}$
$= \dfrac{x + y}{2(x - y)} - \dfrac{x - y}{2(x + y)} + \dfrac{2y^2}{(x + y)(x - y)}$
$= \dfrac{(x + y)^2}{2(x + y)(x - y)} - \dfrac{(x - y)^2}{2(x + y)(x - y)} + \dfrac{4y^2}{2(x + y)(x - y)}$
$= \dfrac{x^2 + 2xy + y^2}{2(x + y)(x - y)} - \dfrac{x^2 - 2xy + y^2}{2(x + y)(x - y)} + \dfrac{4y^2}{2(x + y)(x - y)}$
$= \dfrac{x^2 + 2xy + y^2 - x^2 + 2xy - y^2 + 4y^2}{2(x + y)(x - y)}$
$= \dfrac{4xy + 4y^2}{2(x + y)(x - y)}$
$= \dfrac{4y(x + y)}{2(x + y)(x - y)}$
$= \dfrac{2y}{x - y}$
Bài 2:
a) $\dfrac{1}{k} - \dfrac{1}{k + 2}$
$= \dfrac{k + 2 - k}{k(k + 2)}$
$= \dfrac{2}{k(k + 2)}$
b) $C = \dfrac{2}{k(k + 2)} + \dfrac{2}{(k + 2)(k + 4)} + \dfrac{2}{(k + 4)(k + 6)}$
$= \dfrac{1}{k} - \dfrac{1}{k + 2} + \dfrac{1}{k + 2} - \dfrac{1}{k + 4}+ \dfrac{1}{k + 4} - \dfrac{1}{k + 6}$
$= \dfrac{1}{k} - \dfrac{1}{k + 6}$
$= \dfrac{k + 6 - k}{k(k + 6)}$
$= \dfrac{6}{k(k + 6)}$
c) $M = \dfrac{2}{1 . 3} + \dfrac{2}{3 . 5} + \dfrac{2}{5 . 7} + ... + \dfrac{2}{45 . 47}$
$= \dfrac{1}{1} - \dfrac{1}{3} + \dfrac{1}{3} -\dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + ... + \dfrac{1}{45} - \dfrac{1}{47}$
$= 1 - \bigg(\dfrac{1}{3} - \dfrac{1}{3}\bigg) + \bigg(\dfrac{1}{5} - \dfrac{1}{5}\bigg) + \bigg(\dfrac{1}{7} - \dfrac{1}{7}\bigg) + ... + \bigg(\dfrac{1}{45} - \dfrac{1}{45}\bigg) - \dfrac{1}{47}$
$= 1 - \dfrac{1}{47}$
$= \dfrac{46}{47}$
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