Solve x^2+3x/2x^2-2div3x^2/2x^2+2x*(1-4/x+3)

$$\frac { x ^ { 2 } + 3 x } { 2 x ^ { 2 } - 2 } \div \frac { 3 x ^ { 2 } } { 2 x ^ { 2 } + 2 x } \times ( 1 - \frac { 4 } { x + 3 } )$$

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Short Solution Steps

Divide \frac{x^{2}+3x}{2x^{2}-2} by \frac{3x^{2}}{2x^{2}+2x} by multiplying \frac{x^{2}+3x}{2x^{2}-2} by the reciprocal of \frac{3x^{2}}{2x^{2}+2x}.

$$\frac{\left(x^{2}+3x\right)\left(2x^{2}+2x\right)}{\left(2x^{2}-2\right)\times 3x^{2}}\left(1-\frac{4}{x+3}\right)$$

Factor the expressions that are not already factored in \frac{\left(x^{2}+3x\right)\left(2x^{2}+2x\right)}{\left(2x^{2}-2\right)\times 3x^{2}}.

$$\frac{2\left(x+1\right)\left(x+3\right)x^{2}}{2\times 3\left(x-1\right)\left(x+1\right)x^{2}}\left(1-\frac{4}{x+3}\right)$$

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

$$\frac{x+3}{3\left(x-1\right)}\left(1-\frac{4}{x+3}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+3}{x+3}.

$$\frac{x+3}{3\left(x-1\right)}\left(\frac{x+3}{x+3}-\frac{4}{x+3}\right)$$

Since \frac{x+3}{x+3} and \frac{4}{x+3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x+3}{3\left(x-1\right)}\times \frac{x+3-4}{x+3}$$

Combine like terms in x+3-4.

$$\frac{x+3}{3\left(x-1\right)}\times \frac{x-1}{x+3}$$

Multiply \frac{x+3}{3\left(x-1\right)} times \frac{x-1}{x+3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(x+3\right)\left(x-1\right)}{3\left(x-1\right)\left(x+3\right)}$$

Cancel out \left(x-1\right)\left(x+3\right) in both numerator and denominator.

$$\frac{1}{3}$$

Factor

$$\frac{1}{3} = 0.3333333333333333$$