zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的代数 >

展开 1/(2x+1)+2/(x-1)

解答

展开

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

解答

\frac{5 x + 1}{2 x ^{2} - x - 1}

求解步骤

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

2 x + 1, x - 1

的最小公倍数:

(x - 1) (2 x + 1)

根据最小公倍数调整分式

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

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{x - 1 + 2 ( 2 x + 1 )}{( x - 1 ) ( 2 x + 1 )}

乘开

(x - 1) (2 x + 1) : 2 x^{2} - x - 1

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

乘开

x - 1 + 2 (2 x + 1) : 5 x + 1

= \frac{5 x + 1}{2 x ^{2} - x - 1}

流行的例子

展开 ((6-x)^2)/4

e x p an d \frac{( 6 - x ) ^{2}}{4}

展开 |ax^2+(a+b)x-c|

e x p an d a x^{2} + (a + b) x - c

展开 (3-(e^x)/(10))^2

e x p an d (3 - \frac{e ^{x}}{10})^{2}

展开 ((x,y)-(5,sqrt(2)))*(-3,1)

e x p an d ((x, y) - (5, 2)) \cdot (- 3, 1)

展开 (11(x-24))/2

e x p an d \frac{11 ( x - 24 )}{2}