What is ((2m+n))/((4m^2-n^2)) ?

The solution to ((2m+n))/((4m^2-n^2)) is 1/(2m-n)

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simplify ((2m+n))/((4m^2-n^2))

simplify

\frac{( 2 m + n )}{( 4 m ^{2} - n ^{2} )}

\frac{1}{2 m - n}

Solution steps

\frac{2 m + n}{4 m ^{2} - n ^{2}}

Factor

4 m^{2} - n^{2} : (2 m + n) (2 m - n)

= \frac{2 m + n}{( 2 m + n ) ( 2 m - n )}

Cancel the common factor:

2 m + n

= \frac{1}{2 m - n}

(2 x + 4 x)^{3}

6 p + 18 p^{2} - 4 p^{3}

2 x^{2} - 5 x y

(i 1) x^{s} in x + (sin (x))^{c} os x

(\frac{( 2 x )}{5 + x ^{3}})^{2}

What is ((2m+n))/((4m^2-n^2)) ?
The solution to ((2m+n))/((4m^2-n^2)) is 1/(2m-n)