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Làm các phép tính sau: y / (2x^2 - xy)

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Giải Toán 8 Bài 5: Phép cộng các phân thức đại số

Video Giải Bài 23 trang 46 Toán 8 Tập 1

Bài 23 trang 46 Toán 8 Tập 1:

a) $\frac{y}{2 x^{2} - x y} + \frac{4 x}{y^{2} - 2 x y};$

b) $\frac{1}{x + 2} + \frac{3}{x^{2} - 4} + \frac{x - 14}{(x^{2} + 4 x + 4) (x - 2)};$

c) $\frac{1}{x + 2} + \frac{1}{(x + 2) (4 x + 7)};$

d) $\frac{1}{x + 2} + \frac{1}{(x + 3) (x + 2)} + \frac{1}{(x + 2) (4 x + 7)} .$

Lời giải

a) $\frac{y}{2 x^{2} - x y} + \frac{4 x}{y^{2} - 2 x y}$

$= \frac{y}{x (2 x - y)} + \frac{4 x}{y (y - 2 x)} = \frac{y}{x (2 x - y)} - \frac{4 x}{y (2 x - y)} = \frac{y^{2}}{x y (2 x - y)} - \frac{4 x^{2}}{x y (2 x - y)} = \frac{y^{2} - 4 x^{2}}{x y (2 x - y)} = \frac{y^{2} - {(2 x)}^{2}}{x y (2 x - y)} = \frac{(y - 2 x) (y + 2 x)}{x y (2 x - y)} = \frac{- (2 x - y) (y + 2 x)}{x y (2 x - y)} = \frac{- y - 2 x}{x y}$

b) $\frac{1}{x + 2} + \frac{3}{x^{2} - 4} + \frac{x - 14}{(x^{2} + 4 x + 4) (x - 2)}$

$= \frac{1}{x + 2} + \frac{3}{(x - 2) (x + 2)} + \frac{x - 14}{{(x + 2)}^{2} (x - 2)} = \frac{1 (x + 2) (x - 2)}{{(x + 2)}^{2} (x - 2)} + \frac{3 (x + 2)}{{(x + 2)}^{2} (x - 2)} + \frac{x - 14}{{(x + 2)}^{2} (x - 2)} = \frac{x^{2} - 4}{{(x + 2)}^{2} (x - 2)} + \frac{3 x + 6}{{(x + 2)}^{2} (x - 2)} + \frac{x - 14}{{(x + 2)}^{2} (x - 2)} = \frac{x^{2} - 4 + 3 x + 6 + x - 14}{{(x + 2)}^{2} (x - 2)} = \frac{x^{2} + 4 x - 12}{{(x + 2)}^{2} (x - 2)} = \frac{(x + 6) (x - 2)}{{(x + 2)}^{2} (x - 2)} = \frac{x + 6}{{(x + 2)}^{2}}$

c) $\frac{1}{x + 2} + \frac{1}{(x + 2) (4 x + 7)}$

$= \frac{4 x + 7}{(x + 2) (4 x + 7)} + \frac{1}{(x + 2) (4 x + 7)} = \frac{4 x + 7 + 1}{(x + 2) (4 x + 7)} = \frac{4 x + 8}{(x + 2) (4 x + 7)} = \frac{4 (x + 2)}{(x + 2) (4 x + 7)} = \frac{4}{4 x + 7} .$

d) $\frac{1}{x + 3} + \frac{1}{(x + 3) (x + 2)} + \frac{1}{(x + 2) (4 x + 7)}$

$= \frac{(x + 2) (4 x + 7)}{(x + 3) (x + 2) (4 x + 7)} + \frac{4 x + 7}{(x + 3) (x + 2) (4 x + 7)} + \frac{x + 3}{(x + 3) (x + 2) (4 x + 7)} = \frac{4 x^{2} + 7 x + 8 x + 14}{(x + 3) (x + 2) (4 x + 7)} + \frac{4 x + 7}{(x + 3) (x + 2) (4 x + 7)} + \frac{x + 3}{(x + 3) (x + 2) (4 x + 7)} = \frac{4 x^{2} + 7 x + 8 x + 14 + 4 x + 7 + x + 3}{(x + 3) (x + 2) (4 x + 7)} = \frac{4 x^{2} + 20 x + 24}{(x + 3) (x + 2) (4 x + 7)} = \frac{4 (x^{2} + 5 x + 6)}{(x + 3) (x + 2) (4 x + 7)} = \frac{4 (x + 3) (x + 2)}{(x + 3) (x + 2) (4 x + 7)} = \frac{4}{4 x + 7}$