Solve 2m-8a^2m/1+2a:4am^2-2m^2/2am^2-m

$$\frac { 2 m - 8 a ^ { 2 } m } { 1 + 2 a } : \frac { 4 a m ^ { 2 } - 2 m ^ { 2 } } { 2 a m ^ { 2 } - m }$$

Evaluate

$$1-2am$$

Short Solution Steps

Divide \frac{2m-8a^{2}m}{1+2a} by \frac{4am^{2}-2m^{2}}{2am^{2}-m} by multiplying \frac{2m-8a^{2}m}{1+2a} by the reciprocal of \frac{4am^{2}-2m^{2}}{2am^{2}-m}.

$$\frac{\left(2m-8a^{2}m\right)\left(2am^{2}-m\right)}{\left(1+2a\right)\left(4am^{2}-2m^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{2\left(-2a-1\right)\left(2a-1\right)m^{2}\left(2am-1\right)}{2\left(2a-1\right)\left(2a+1\right)m^{2}}$$

Extract the negative sign in -1-2a.

$$\frac{-2\left(2a-1\right)\left(2a+1\right)m^{2}\left(2am-1\right)}{2\left(2a-1\right)\left(2a+1\right)m^{2}}$$

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

$$-\left(2am-1\right)$$

Expand the expression.

$$-2am+1$$

Expand

$$1-2am$$

Short Solution Steps

Divide \frac{2m-8a^{2}m}{1+2a} by \frac{4am^{2}-2m^{2}}{2am^{2}-m} by multiplying \frac{2m-8a^{2}m}{1+2a} by the reciprocal of \frac{4am^{2}-2m^{2}}{2am^{2}-m}.

$$\frac{\left(2m-8a^{2}m\right)\left(2am^{2}-m\right)}{\left(1+2a\right)\left(4am^{2}-2m^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{2\left(-2a-1\right)\left(2a-1\right)m^{2}\left(2am-1\right)}{2\left(2a-1\right)\left(2a+1\right)m^{2}}$$

Extract the negative sign in -1-2a.

$$\frac{-2\left(2a-1\right)\left(2a+1\right)m^{2}\left(2am-1\right)}{2\left(2a-1\right)\left(2a+1\right)m^{2}}$$

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

$$-\left(2am-1\right)$$

Expand the expression.

$$-2am+1$$