Question

$$\frac{ \sqrt{ { x }^{ 4 } - { x }^{ 3 } } }{ 1-2x } =y$$

Answer

$$y=(x^(3/2)*sqrt(x-1))/(1-2*x)$$

Solution


Factor out the common term \({x}^{3}\).
\[\frac{\sqrt{{x}^{3}(x-1)}}{1-2x}=y\]
Use this rule: \(\sqrt{ab}=\sqrt{a}\sqrt{b}\).
\[\frac{\sqrt{{x}^{3}}\sqrt{x-1}}{1-2x}=y\]
Simplify  \(\sqrt{{x}^{3}}\)  to  \({({x}^{3})}^{\frac{1}{2}}\).
\[\frac{{({x}^{3})}^{\frac{1}{2}}\sqrt{x-1}}{1-2x}=y\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{{x}^{\frac{3}{2}}\sqrt{x-1}}{1-2x}=y\]
Switch sides.
\[y=\frac{{x}^{\frac{3}{2}}\sqrt{x-1}}{1-2x}\]