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

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

Evaluate

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

Short Solution Steps

The opposite of -1 is 1.

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

Add -2 and 1 to get -1.

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

Fraction \frac{-2}{-1} can be simplified to 2 by removing the negative sign from both the numerator and the denominator.

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

Factor x^{2}-2xy+y^{2}.

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

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

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

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

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

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

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

Combine like terms in 2x^{2}-4xy+2y^{2}+2xy.

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

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

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

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

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

Do the multiplications in 2x-\left(x+y\right).

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

Combine like terms in 2x-x-y.

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

Multiply \frac{2x^{2}+2y^{2}-2xy}{\left(x-y\right)^{2}} times \frac{x-y}{x+y} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out x-y in both numerator and denominator.

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

Consider \left(x+y\right)\left(x-y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand

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

Short Solution Steps

The opposite of -1 is 1.

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

Add -2 and 1 to get -1.

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

Fraction \frac{-2}{-1} can be simplified to 2 by removing the negative sign from both the numerator and the denominator.

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

Factor x^{2}-2xy+y^{2}.

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

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

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

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

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

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

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

Combine like terms in 2x^{2}-4xy+2y^{2}+2xy.

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

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

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

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

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

Do the multiplications in 2x-\left(x+y\right).

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

Combine like terms in 2x-x-y.

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

Multiply \frac{2x^{2}+2y^{2}-2xy}{\left(x-y\right)^{2}} times \frac{x-y}{x+y} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out x-y in both numerator and denominator.

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

Consider \left(x+y\right)\left(x-y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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