Solve A+z/A^2+A-1/A+A+1/-A^2-2A-1=

$$\frac { A + z } { A ^ { 2 } + A } - \frac { 1 } { A } + \frac { A + 1 } { - A ^ { 2 } - 2 A - 1 } =$$

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{A+1}{-A^{2}-2A-1}.

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

Extract the negative sign in A+1. Extract the negative sign in -1-A.

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

Cancel out A+1 in both numerator and denominator.

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

Multiply -1 and -1 to get 1.

$$\frac{A+z}{A^{2}+A}-\frac{1}{A}+\frac{1}{-A-1}$$

Factor A^{2}+A.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of A\left(A+1\right) and A is A\left(A+1\right). Multiply \frac{1}{A} times \frac{A+1}{A+1}.

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

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

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

Do the multiplications in A+z-\left(A+1\right).

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

Combine like terms in A+z-A-1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of A\left(A+1\right) and -A-1 is A\left(A+1\right). Multiply \frac{1}{-A-1} times \frac{-A}{-A}.

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

Since \frac{z-1}{A\left(A+1\right)} and \frac{-A}{A\left(A+1\right)} have the same denominator, add them by adding their numerators.

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

Expand A\left(A+1\right).

$$\frac{z-1-A}{A^{2}+A}$$

Expand

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{A+1}{-A^{2}-2A-1}.

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

Extract the negative sign in A+1. Extract the negative sign in -1-A.

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

Cancel out A+1 in both numerator and denominator.

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

Multiply -1 and -1 to get 1.

$$\frac{A+z}{A^{2}+A}-\frac{1}{A}+\frac{1}{-A-1}$$

Factor A^{2}+A.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of A\left(A+1\right) and A is A\left(A+1\right). Multiply \frac{1}{A} times \frac{A+1}{A+1}.

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

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

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

Do the multiplications in A+z-\left(A+1\right).

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

Combine like terms in A+z-A-1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of A\left(A+1\right) and -A-1 is A\left(A+1\right). Multiply \frac{1}{-A-1} times \frac{-A}{-A}.

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

Since \frac{z-1}{A\left(A+1\right)} and \frac{-A}{A\left(A+1\right)} have the same denominator, add them by adding their numerators.

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

Expand A\left(A+1\right).

$$\frac{z-1-A}{A^{2}+A}$$