What is expand 1/(3x^2-3)-2/(2x+2)-(x+5)/(x+1) ?

The solution to expand 1/(3x^2-3)-2/(2x+2)-(x+5)/(x+1) is (-3x^2-15x+19)/(3x^2-3)

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expand 1/(3x^2-3)-2/(2x+2)-(x+5)/(x+1)

expand

\frac{1}{3 x ^{2} - 3} - \frac{2}{2 x + 2} - \frac{x + 5}{x + 1}

\frac{- 3 x ^{2} - 15 x + 19}{3 x ^{2} - 3}

Solution steps

\frac{1}{3 x ^{2} - 3} - \frac{2}{2 x + 2} - \frac{x + 5}{x + 1}

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

= \frac{1}{3 x ^{2} - 3} - \frac{1}{x + 1} - \frac{x + 5}{x + 1}

Combine the fractions

- \frac{1}{x + 1} - \frac{x + 5}{x + 1} : \frac{- 1 - ( x + 5 )}{x + 1}

= \frac{1}{3 x ^{2} - 3} + \frac{- ( x + 5 ) - 1}{x + 1}

Expand

- 1 - (x + 5) : - x - 6

= \frac{1}{3 x ^{2} - 3} + \frac{- x - 6}{x + 1}

Factor

3 x^{2} - 3 : 3 (x + 1) (x - 1)

= \frac{1}{3 ( x + 1 ) ( x - 1 )} + \frac{- x - 6}{x + 1}

Least Common Multiplier of

3 (x + 1) (x - 1), x + 1 : 3 (x + 1) (x - 1)

Adjust Fractions based on the LCM

= \frac{1}{3 ( x + 1 ) ( x - 1 )} + \frac{( - x - 6 ) \cdot 3 ( x - 1 )}{( x + 1 ) \cdot 3 ( x - 1 )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + ( - x - 6 ) \cdot 3 ( x - 1 )}{3 ( x + 1 ) ( x - 1 )}

Expand

3 (x + 1) (x - 1) : 3 x^{2} - 3

= \frac{1 + 3 ( - x - 6 ) ( x - 1 )}{3 x ^{2} - 3}

Expand

1 + (- x - 6) \cdot 3 (x - 1) : - 3 x^{2} - 15 x + 19

= \frac{- 3 x ^{2} - 15 x + 19}{3 x ^{2} - 3}

\frac{2}{( s + 4 ) ( s + 1 )}

(2 r cos (θ) - r sin (θ)) r

\frac{2 s ^{2} + 2 s + 1}{s ^{3} ( s + 1 )}

\frac{( e ^{\frac{y}{t}} )}{( e ^{\frac{y}{t}} ) ^{2}}

(x + z) (sin (t))

What is expand 1/(3x^2-3)-2/(2x+2)-(x+5)/(x+1) ?
The solution to expand 1/(3x^2-3)-2/(2x+2)-(x+5)/(x+1) is (-3x^2-15x+19)/(3x^2-3)