What is expand (1-x)/(2-2x)((1-2x)/(2-2x))^2 ?

The solution to expand (1-x)/(2-2x)((1-2x)/(2-2x))^2 is (1-4x+4x^2)/(8-16x+8x^2)

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expand (1-x)/(2-2x)((1-2x)/(2-2x))^2

expand

\frac{1 - x}{2 - 2 x} (\frac{1 - 2 x}{2 - 2 x})^{2}

\frac{1 - 4 x + 4 x ^{2}}{8 - 16 x + 8 x ^{2}}

Solution steps

\frac{1 - x}{2 - 2 x} (\frac{1 - 2 x}{2 - 2 x})^{2}

\frac{1 - x}{2 - 2 x} = \frac{1}{2}

= \frac{1}{2} (\frac{- 2 x + 1}{- 2 x + 2})^{2}

(\frac{1 - 2 x}{2 - 2 x})^{2} = \frac{1 - 4 x + 4 x ^{2}}{4 - 8 x + 4 x ^{2}}

= \frac{1}{2} \cdot \frac{4 x ^{2} - 4 x + 1}{4 x ^{2} - 8 x + 4}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot ( 1 - 4 x + 4 x ^{2} )}{2 ( 4 - 8 x + 4 x ^{2} )}

1 \cdot (1 - 4 x + 4 x^{2}) = 1 - 4 x + 4 x^{2}

= \frac{1 - 4 x + 4 x ^{2}}{2 ( 4 x ^{2} - 8 x + 4 )}

Expand

2 (4 - 8 x + 4 x^{2}) : 8 - 16 x + 8 x^{2}

= \frac{1 - 4 x + 4 x ^{2}}{8 - 16 x + 8 x ^{2}}

- 5 (- v + 6 w + (- \frac{1}{2}) w)

\frac{1}{( 2 x - 3 ) ( 2 x + 3 )}

\frac{30 ( - k e ^{- k t} t - e ^{- k t} )}{k}

\frac{( 50 + b ) ^{2}}{4}

f^{'} (2) (X^{2} - 1) (2 X + 2)

What is expand (1-x)/(2-2x)((1-2x)/(2-2x))^2 ?
The solution to expand (1-x)/(2-2x)((1-2x)/(2-2x))^2 is (1-4x+4x^2)/(8-16x+8x^2)