What is factor (\frac{8x)/(x^2-1)}{(10x)/(x+1)} ?

The solution to factor (\frac{8x)/(x^2-1)}{(10x)/(x+1)} is 4/(5(x-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

factor (\frac{8x)/(x^2-1)}{(10x)/(x+1)}

factor

\frac{\frac{8 x}{x ^{2} - 1}}{\frac{10 x}{x + 1}}

\frac{4}{5 ( x - 1 )}

Solution steps

\frac{\frac{8 x}{x ^{2} - 1}}{\frac{10 x}{x + 1}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{8 x ( x + 1 )}{( x ^{2} - 1 ) \cdot 10 x}

Cancel the common factor:

x

= \frac{8 ( x + 1 )}{( x ^{2} - 1 ) \cdot 10}

Cancel the common factor:

2

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

Factor

5 (x^{2} - 1) : 5 (x + 1) (x - 1)

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

Cancel the common factor:

x + 1

= \frac{4}{5 ( x - 1 )}

- 16 t^{2} + 36 t

\frac{3 π}{4}

\frac{10}{3 ( 4 ) ^{\frac{7}{2}}}

5 x^{2} - 2 x + 5 = 0

8 \div 2 (2 + 2)

What is factor (\frac{8x)/(x^2-1)}{(10x)/(x+1)} ?
The solution to factor (\frac{8x)/(x^2-1)}{(10x)/(x+1)} is 4/(5(x-1))