Solve x^2+2x+1/x^2+2xdiv(1-1/x+2)

$$\frac { x ^ { 2 } + 2 x + 1 } { x ^ { 2 } + 2 x } \div ( 1 - \frac { 1 } { x + 2 } )$$

Evaluate

$$1+\frac{1}{x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+2}{x+2}.

$$\frac{\frac{x^{2}+2x+1}{x^{2}+2x}}{\frac{x+2}{x+2}-\frac{1}{x+2}}$$

Since \frac{x+2}{x+2} and \frac{1}{x+2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x^{2}+2x+1}{x^{2}+2x}}{\frac{x+2-1}{x+2}}$$

Combine like terms in x+2-1.

$$\frac{\frac{x^{2}+2x+1}{x^{2}+2x}}{\frac{x+1}{x+2}}$$

Divide \frac{x^{2}+2x+1}{x^{2}+2x} by \frac{x+1}{x+2} by multiplying \frac{x^{2}+2x+1}{x^{2}+2x} by the reciprocal of \frac{x+1}{x+2}.

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

Factor the expressions that are not already factored.

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

Cancel out \left(x+1\right)\left(x+2\right) in both numerator and denominator.

$$\frac{x+1}{x}$$

Expand

$$1+\frac{1}{x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+2}{x+2}.

$$\frac{\frac{x^{2}+2x+1}{x^{2}+2x}}{\frac{x+2}{x+2}-\frac{1}{x+2}}$$

Since \frac{x+2}{x+2} and \frac{1}{x+2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x^{2}+2x+1}{x^{2}+2x}}{\frac{x+2-1}{x+2}}$$

Combine like terms in x+2-1.

$$\frac{\frac{x^{2}+2x+1}{x^{2}+2x}}{\frac{x+1}{x+2}}$$

Divide \frac{x^{2}+2x+1}{x^{2}+2x} by \frac{x+1}{x+2} by multiplying \frac{x^{2}+2x+1}{x^{2}+2x} by the reciprocal of \frac{x+1}{x+2}.

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

Factor the expressions that are not already factored.

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

Cancel out \left(x+1\right)\left(x+2\right) in both numerator and denominator.

$$\frac{x+1}{x}$$