Step 1

Given:

\(\displaystyle{\frac{{{3}}}{{{2}{x}+{5}}}}+{\frac{{{4}}}{{{2}{x}-{5}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}\)

Now we simplify the above equation as,

\(\displaystyle\Rightarrow{\frac{{{3}{\left({2}{x}-{5}\right)}+{4}{\left({2}{x}+{5}\right)}}}{{{4}{x}^{{{2}}}-{25}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}\)

\(\displaystyle\Rightarrow{\frac{{{6}{x}-{15}+{8}{x}+{20}}}{{{4}{x}^{{{2}}}-{25}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}\)

\(\displaystyle\Rightarrow{14}{x}+{5}={14}{x}+{3}\)

\(\displaystyle\Rightarrow{5}={3}\)

Step 2

Which is not possible.

Since,

\(\displaystyle{5}\ne{3}\)

Therefore equation has no solution

Given:

\(\displaystyle{\frac{{{3}}}{{{2}{x}+{5}}}}+{\frac{{{4}}}{{{2}{x}-{5}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}\)

Now we simplify the above equation as,

\(\displaystyle\Rightarrow{\frac{{{3}{\left({2}{x}-{5}\right)}+{4}{\left({2}{x}+{5}\right)}}}{{{4}{x}^{{{2}}}-{25}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}\)

\(\displaystyle\Rightarrow{\frac{{{6}{x}-{15}+{8}{x}+{20}}}{{{4}{x}^{{{2}}}-{25}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}\)

\(\displaystyle\Rightarrow{14}{x}+{5}={14}{x}+{3}\)

\(\displaystyle\Rightarrow{5}={3}\)

Step 2

Which is not possible.

Since,

\(\displaystyle{5}\ne{3}\)

Therefore equation has no solution