Solve 1+6/x+9/x^2=

$$1 + \frac { 6 } { x } + \frac { 9 } { x ^ { 2 } } =$$

Evaluate

$$\left(\frac{x+3}{x}\right)^{2}$$

Short Solution Steps

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

$$\frac{x}{x}+\frac{6}{x}+\frac{9}{x^{2}}$$

Since \frac{x}{x} and \frac{6}{x} have the same denominator, add them by adding their numerators.

$$\frac{x+6}{x}+\frac{9}{x^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and x^{2} is x^{2}. Multiply \frac{x+6}{x} times \frac{x}{x}.

$$\frac{\left(x+6\right)x}{x^{2}}+\frac{9}{x^{2}}$$

Since \frac{\left(x+6\right)x}{x^{2}} and \frac{9}{x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x+6\right)x+9}{x^{2}}$$

Do the multiplications in \left(x+6\right)x+9.

$$\frac{x^{2}+6x+9}{x^{2}}$$

Factor

$$\frac{\left(x+3\right)^{2}}{x^{2}}$$