Write as a single fraction in its simplest form $\dfrac{3}{(1-2h)^2}+\dfrac{8}{2h-1}$ . Answer ...

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Worked solution

1. Rewrite the denominator so both fractions are in terms of $2h-1$.

Since $1-2h=-(2h-1)$ ,
$(1-2h)^2=(2h-1)^2.$
So
$\frac{3}{(1-2h)^2}+\frac{8}{2h-1}=\frac{3}{(2h-1)^2}+\frac{8}{2h-1}.$

2. Express both terms over the common denominator $(2h-1)^2$.

\frac{3}{(2h-1)^2}+\frac{8}{2h-1} =\frac{3}{(2h-1)^2}+\frac{8(2h-1)}{(2h-1)^2}.

3. Combine the numerators and simplify.

\frac{3}{(2h-1)^2}+\frac{8(2h-1)}{(2h-1)^2} =\frac{3+8(2h-1)}{(2h-1)^2}

=\frac{3+16h-8}{(2h-1)^2} =\frac{16h-5}{(2h-1)^2}.

Answer

$\dfrac{16h-5}{(2h-1)^2}$

AI analysis — may contain mistakes.

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