Solve 1/2x-1/x+y*(x^2-y^2+x+y/2x)

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

Evaluate

$$y-x$$

Short Solution Steps

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

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

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

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

Do the multiplications in \left(x^{2}-y^{2}\right)\times 2x+x+y.

$$\frac{1}{2x}-\frac{1}{x+y}\times \frac{2x^{3}-2y^{2}x+x+y}{2x}$$

Multiply \frac{1}{x+y} times \frac{2x^{3}-2y^{2}x+x+y}{2x} by multiplying numerator times numerator and denominator times denominator.

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

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

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

Extract the negative sign in -x-y.

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

Cancel out x+y in both numerator and denominator.

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

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

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

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

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

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

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

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

$$\frac{2x\left(-x+y\right)}{2x}$$

Cancel out 2x in both numerator and denominator.

$$-x+y$$

Expand

$$y-x$$

Short Solution Steps

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

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

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

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

Do the multiplications in \left(x^{2}-y^{2}\right)\times 2x+x+y.

$$\frac{1}{2x}-\frac{1}{x+y}\times \frac{2x^{3}-2y^{2}x+x+y}{2x}$$

Multiply \frac{1}{x+y} times \frac{2x^{3}-2y^{2}x+x+y}{2x} by multiplying numerator times numerator and denominator times denominator.

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

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

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

Extract the negative sign in -x-y.

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

Cancel out x+y in both numerator and denominator.

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

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

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

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

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

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

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

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

$$\frac{2x\left(-x+y\right)}{2x}$$

Cancel out 2x in both numerator and denominator.

$$-x+y$$