展开 ((2x+2b)^2)/(2x)

解答

展开

\frac{( 2 x + 2 b ) ^{2}}{2 x}

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

求解步骤

\frac{( 2 x + 2 b ) ^{2}}{2 x}

(2 x + 2 b)^{2} = 4 x^{2} + 8 b x + 4 b^{2}

= \frac{4 x ^{2} + 8 b x + 4 b ^{2}}{2 x}

分解

4 x^{2} + 8 b x + 4 b^{2} : 4 (x^{2} + 2 b x + b^{2})

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

数字相除:

\frac{4}{2} = 2

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

乘开

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

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

使用分式法则:

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

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

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

消掉

\frac{2 x ^{2}}{x} : 2 x

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

消掉

\frac{4 b x}{x} : 4 b

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

e x p an d \frac{d x}{( 3 - 5 x ) ^{2}}

e x p an d \frac{lo g _{2} ( n ) ( 3 + lo g _{2} ( n ) )}{2}

e x p an d \frac{4 - c ^{2} ( 3 + c ^{2} + 5 )}{8 + c ^{2}}

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

e x p an d c_{2} (e^{- 5 x} - 5 e^{- 5 x} x)