展开 (2-x)/(x^2+4x+4)+1/(x+2)

解答

展开

\frac{2 - x}{x ^{2} + 4 x + 4} + \frac{1}{x + 2}

\frac{4}{x ^{2} + 4 x + 4}

求解步骤

\frac{2 - x}{x ^{2} + 4 x + 4} + \frac{1}{x + 2}

分解

x^{2} + 4 x + 4 : (x + 2)^{2}

= \frac{2 - x}{( x + 2 ) ^{2}} + \frac{1}{x + 2}

(x + 2)^{2}, x + 2

的最小公倍数:

(x + 2)^{2}

根据最小公倍数调整分式

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

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

2 - x + x + 2 = 4

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

(x + 2)^{2} = x^{2} + 4 x + 4

= \frac{4}{x ^{2} + 4 x + 4}

e x p an d \frac{2 n + 1}{n ( n + 1 )}

e x p an d (2 cos (2 x) + 2 sin (2 x)) (sin (2 x) + cos (2 x))

s im pl i f y (2 x + x)^{2}

\frac{d}{d t} ((x) (32 (t - sin (t))))

e x p an d (1 + 3 i, - 2 i, 2 + 3 i) \cdot (2, 1 + 2 i, - 1 + i)