What is 1/t-1/(t^2+t) ?

The solution to 1/t-1/(t^2+t) is 1/(t+1)

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simplify 1/t-1/(t^2+t)

simplify

\frac{1}{t} - \frac{1}{t ^{2} + t}

\frac{1}{t + 1}

Solution steps

\frac{1}{t} - \frac{1}{t ^{2} + t}

Least Common Multiplier of

t, t^{2} + t : t (t + 1)

Adjust Fractions based on the LCM

= \frac{t + 1}{t ( t + 1 )} - \frac{1}{t ( t + 1 )}

Apply the fraction rule:

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

= \frac{t + 1 - 1}{t ( t + 1 )}

Subtract the numbers:

1 - 1 = 0

= \frac{t}{t ( t + 1 )}

Cancel the common factor:

t

= \frac{1}{t + 1}

∣ a + b ∣^{2} - ∣ a - b ∣^{2}

32 y^{5} - 1

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

v (t)

\frac{2}{5} \cdot \frac{15}{4}