<s>[INST] Radical equation $\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt[4]{4x-3}}=\frac{2}{3}$ I am stuck in the following difficult radical equation:
$$
\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt[4]{4x-3}}=\frac{2}{3}
$$
Attemption. The left hand side of the equation is a decrease function. Therefore, the equation has a unique solution. An approximation for the solution is 8.74874 (using Matlab).
Thank you for all kind help hint.
 [/INST] Hint: $u=2x-1$ and $v=4x-3$ 
Rewrite the expression:
$$
\frac{1}{\sqrt{u}}+\frac{1}{\sqrt[4]{v}}=\frac{2}{3} \implies \frac{\sqrt[4]{v}+\sqrt{u}}{\sqrt{u}\sqrt[4]{v}}=\frac{2}{3}
$$
$$
\implies \sqrt[4]{v}+\sqrt{u}=\frac 23\sqrt{u}\sqrt[4]{v}
$$
$$
\implies \sqrt{u}=\left(\frac 23\sqrt{u}-1\right)\sqrt[4]{v} \implies \frac{\sqrt{u}}{\frac 23\sqrt{u}-1}=\sqrt[4]{v} 
$$
Now, raise the equation to the fourth power:
$$
\implies \frac{u^2}{(\frac 23\sqrt{u}-1)^4}=v 
$$
Now isolate $\sqrt{u}$ on one side of the equation and raise everything to the second power. The resulting polynomial in $x$ after inverting the substitutions might not be solvable by an analytical expression.
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