What is expand (2x+1)/(4(x^2+x)^{5/4)} ?

The solution to expand (2x+1)/(4(x^2+x)^{5/4)} is x/(2(x^2+x)^{5/4)}+1/(4(x^2+x)^{5/4)}

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expand (2x+1)/(4(x^2+x)^{5/4)}

expand

\frac{2 x + 1}{4 ( x ^{2} + x ) ^{\frac{5}{4}}}

\frac{x}{2 ( x ^{2} + x ) ^{\frac{5}{4}}} + \frac{1}{4 ( x ^{2} + x ) ^{\frac{5}{4}}}

Solution steps

\frac{2 x + 1}{4 ( x ^{2} + x ) ^{\frac{5}{4}}}

Apply the fraction rule:

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

\frac{2 x + 1}{4 ( x ^{2} + x ) ^{\frac{5}{4}}} = \frac{2 x}{4 ( x ^{2} + x ) ^{\frac{5}{4}}} + \frac{1}{4 ( x ^{2} + x ) ^{\frac{5}{4}}}

= \frac{2 x}{4 ( x ^{2} + x ) ^{\frac{5}{4}}} + \frac{1}{4 ( x ^{2} + x ) ^{\frac{5}{4}}}

Cancel

\frac{2 x}{4 ( x ^{2} + x ) ^{\frac{5}{4}}} : \frac{x}{2 ( x ^{2} + x ) ^{\frac{5}{4}}}

= \frac{x}{2 ( x ^{2} + x ) ^{\frac{5}{4}}} + \frac{1}{4 ( x ^{2} + x ) ^{\frac{5}{4}}}

(- 67 y - 97)^{2}

(3 X - 4 X^{2}) (6 X + 1)^{''}

- \frac{2 x ( x + 1 ) ^{2}}{3}

3 - sin (2 x^{2} - 3 x - 1) (4 x - 3)

What is expand (2x+1)/(4(x^2+x)^{5/4)} ?
The solution to expand (2x+1)/(4(x^2+x)^{5/4)} is x/(2(x^2+x)^{5/4)}+1/(4(x^2+x)^{5/4)}