Solve [1-1/(1+frac{1{m})(1-1/m)}]:1/1-m

$$[ 1 - \frac { 1 } { ( 1 + \frac { 1 } { m } ) ( 1 - \frac { 1 } { m } ) } ] : \frac { 1 } { 1 - m }$$

Evaluate

$$\frac{1}{m+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{m}{m}.

$$\frac{1-\frac{1}{\left(\frac{m}{m}+\frac{1}{m}\right)\left(1-\frac{1}{m}\right)}}{\frac{1}{1-m}}$$

Since \frac{m}{m} and \frac{1}{m} have the same denominator, add them by adding their numerators.

$$\frac{1-\frac{1}{\frac{m+1}{m}\left(1-\frac{1}{m}\right)}}{\frac{1}{1-m}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{m}{m}.

$$\frac{1-\frac{1}{\frac{m+1}{m}\left(\frac{m}{m}-\frac{1}{m}\right)}}{\frac{1}{1-m}}$$

Since \frac{m}{m} and \frac{1}{m} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-\frac{1}{\frac{m+1}{m}\times \frac{m-1}{m}}}{\frac{1}{1-m}}$$

Multiply \frac{m+1}{m} times \frac{m-1}{m} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1-\frac{1}{\frac{\left(m+1\right)\left(m-1\right)}{mm}}}{\frac{1}{1-m}}$$

Divide 1 by \frac{\left(m+1\right)\left(m-1\right)}{mm} by multiplying 1 by the reciprocal of \frac{\left(m+1\right)\left(m-1\right)}{mm}.

$$\frac{1-\frac{mm}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Multiply m and m to get m^{2}.

$$\frac{1-\frac{m^{2}}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(m+1\right)\left(m-1\right)}{\left(m+1\right)\left(m-1\right)}.

$$\frac{\frac{\left(m+1\right)\left(m-1\right)}{\left(m+1\right)\left(m-1\right)}-\frac{m^{2}}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Since \frac{\left(m+1\right)\left(m-1\right)}{\left(m+1\right)\left(m-1\right)} and \frac{m^{2}}{\left(m+1\right)\left(m-1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\left(m+1\right)\left(m-1\right)-m^{2}}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Do the multiplications in \left(m+1\right)\left(m-1\right)-m^{2}.

$$\frac{\frac{m^{2}-m+m-1-m^{2}}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Combine like terms in m^{2}-m+m-1-m^{2}.

$$\frac{\frac{-1}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Divide \frac{-1}{\left(m+1\right)\left(m-1\right)} by \frac{1}{1-m} by multiplying \frac{-1}{\left(m+1\right)\left(m-1\right)} by the reciprocal of \frac{1}{1-m}.

$$\frac{-\left(1-m\right)}{\left(m+1\right)\left(m-1\right)}$$

Extract the negative sign in 1-m.

$$\frac{-\left(-1\right)\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}$$

Cancel out m-1 in both numerator and denominator.

$$\frac{-\left(-1\right)}{m+1}$$

Multiply -1 and -1 to get 1.

$$\frac{1}{m+1}$$

Expand

$$\frac{1}{m+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{m}{m}.

$$\frac{1-\frac{1}{\left(\frac{m}{m}+\frac{1}{m}\right)\left(1-\frac{1}{m}\right)}}{\frac{1}{1-m}}$$

Since \frac{m}{m} and \frac{1}{m} have the same denominator, add them by adding their numerators.

$$\frac{1-\frac{1}{\frac{m+1}{m}\left(1-\frac{1}{m}\right)}}{\frac{1}{1-m}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{m}{m}.

$$\frac{1-\frac{1}{\frac{m+1}{m}\left(\frac{m}{m}-\frac{1}{m}\right)}}{\frac{1}{1-m}}$$

Since \frac{m}{m} and \frac{1}{m} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-\frac{1}{\frac{m+1}{m}\times \frac{m-1}{m}}}{\frac{1}{1-m}}$$

Multiply \frac{m+1}{m} times \frac{m-1}{m} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1-\frac{1}{\frac{\left(m+1\right)\left(m-1\right)}{mm}}}{\frac{1}{1-m}}$$

Divide 1 by \frac{\left(m+1\right)\left(m-1\right)}{mm} by multiplying 1 by the reciprocal of \frac{\left(m+1\right)\left(m-1\right)}{mm}.

$$\frac{1-\frac{mm}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Multiply m and m to get m^{2}.

$$\frac{1-\frac{m^{2}}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(m+1\right)\left(m-1\right)}{\left(m+1\right)\left(m-1\right)}.

$$\frac{\frac{\left(m+1\right)\left(m-1\right)}{\left(m+1\right)\left(m-1\right)}-\frac{m^{2}}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Since \frac{\left(m+1\right)\left(m-1\right)}{\left(m+1\right)\left(m-1\right)} and \frac{m^{2}}{\left(m+1\right)\left(m-1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\left(m+1\right)\left(m-1\right)-m^{2}}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Do the multiplications in \left(m+1\right)\left(m-1\right)-m^{2}.

$$\frac{\frac{m^{2}-m+m-1-m^{2}}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Combine like terms in m^{2}-m+m-1-m^{2}.

$$\frac{\frac{-1}{\left(m+1\right)\left(m-1\right)}}{\frac{1}{1-m}}$$

Divide \frac{-1}{\left(m+1\right)\left(m-1\right)} by \frac{1}{1-m} by multiplying \frac{-1}{\left(m+1\right)\left(m-1\right)} by the reciprocal of \frac{1}{1-m}.

$$\frac{-\left(1-m\right)}{\left(m+1\right)\left(m-1\right)}$$

Extract the negative sign in 1-m.

$$\frac{-\left(-1\right)\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}$$

Cancel out m-1 in both numerator and denominator.

$$\frac{-\left(-1\right)}{m+1}$$

Multiply -1 and -1 to get 1.

$$\frac{1}{m+1}$$