√(x + 7) - √(6 - x) = 1 add √(6 - x) to both sides to separate the radicals
√(x + 7) = 1 + √(6 - x) square both sides
x + 7 = 1 + 2 √(6 - x) + (6 -x) rearrange to get theradical on one side
2x =2 √(6 - x) divide both sides by 2
x = √(6 - x) square both sides, again
x^2 = 6 - x rearrange
x^2 + x - 6 = 0 factor
(x + 3) ( x - 2) = 0
And setting each factor to 0, we have that x = -3 or x = 2 ......we must check these in the original problem, as these types of equations often lead to "extraneous" solutions [due to the squaring of both sides]
Check -3 ... [√(-3 + 7) - √(6 - (-3)) = 1 ] = [2 - 3 = 1 ] is untrue
Check 2 ... [ √(2 + 7) - √(6 - 2) = 1 ] = [ 3 - 2 ] = 1 which is true
So x = 2 is the only solution.......here's a graph of both sides of the original problem.....
https://www.desmos.com/calculator/rmqhq2arba
Notice that the only solution occurs at (2, 1)
I assume we have....
1 / [2n^2] + 5 / [2n] = n - 2 /[n^2] multiply through by 2n^2....this gives...
1 + 5n = 2n^3 - 4 rearrange
2n^3 - 5n - 5 = 0 this isn't factorable.....the only "real" solution occurs at about n = 1.946
See the graph here......https://www.desmos.com/calculator/q449wlwqlg
$${\mathtt{2}}\left[{{n}}^{{\mathtt{3}}}\right]{\mathtt{\,-\,}}{\mathtt{5}}{n}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\frac{{\mathtt{5}}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)\\
{\mathtt{n}} = {\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{n}} = {\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\mathtt{\,-\,}}{\mathtt{0.972\: \!755\: \!103\: \!270\: \!916\: \!1}}{\mathtt{\,-\,}}{\mathtt{0.582\: \!028\: \!756\: \!006\: \!450\: \!1}}{i}\\
{\mathtt{n}} = {\mathtt{\,-\,}}{\mathtt{0.972\: \!755\: \!103\: \!270\: \!916\: \!1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.582\: \!028\: \!756\: \!006\: \!450\: \!1}}{i}\\
{\mathtt{n}} = {\mathtt{1.945\: \!510\: \!206\: \!541\: \!832\: \!2}}\\
\end{array} \right\}$$