展开 (f(x^2-5+h)-x^2-5)/h

解答

展开

\frac{f ( x ^{2} - 5 + h ) - x ^{2} - 5}{h}

\frac{f x ^{2}}{h} - \frac{5 f}{h} + f - \frac{x ^{2}}{h} - \frac{5}{h}

求解步骤

\frac{f ( x ^{2} - 5 + h ) - x ^{2} - 5}{h}

乘开

f (x^{2} - 5 + h) - x^{2} - 5 : f x^{2} - 5 f + f h - x^{2} - 5

= \frac{f x ^{2} - 5 f + f h - x ^{2} - 5}{h}

使用分式法则:

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

\frac{f x ^{2} - 5 f + f h - x ^{2} - 5}{h} = \frac{f x ^{2}}{h} - \frac{5 f}{h} + \frac{f h}{h} - \frac{x ^{2}}{h} - \frac{5}{h}

= \frac{f x ^{2}}{h} - \frac{5 f}{h} + \frac{f h}{h} - \frac{x ^{2}}{h} - \frac{5}{h}

消掉

\frac{f h}{h} : f

= \frac{f x ^{2}}{h} - \frac{5 f}{h} + f - \frac{x ^{2}}{h} - \frac{5}{h}

e x p an d ln (t) (t ln (t) - t)

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

e x p an d \frac{2 x ^{2} + x + 8}{( 2 x - 1 ) ( x + 3 )}

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

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