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I am terribly stuck in the following questions about inequalities. Any idea how to solve these?

\begin{aligned}\frac{1+(\log_ax)^2}{1+\log_ax}>1,\space0<a<1\end{aligned} \begin{aligned}\sqrt{2-\sqrt{3+x}}<\sqrt{4+x}\end{aligned} \begin{aligned}\log_\sqrt{2x^2-7x+6}\left(\frac{x}{3}\right)>0\end{aligned}

The given answers are \begin{aligned}0<x<a,1<x<\frac{1}{a}\end{aligned} \begin{aligned}-\frac{3+\sqrt5}{2}<x\leq1\end{aligned} \begin{aligned}1<x<\frac{3}{2},2<x<\frac{5}{2},x>3\end{aligned}

Thanks in advance...

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HINT :

For the first, separate it into two cases as $$1+\log_ax\gt 0\ \ \text{or}\ \ 1+\log_ax\lt 0.$$

For the second, we have $$2-\sqrt{3+x}\ge 0\ \ \text{and}\ \ 3+x\ge0\ \ \text{and}\ \ 4+x\ge 0$$ and we have $$\sqrt{2-\sqrt{3+x}}\lt\sqrt{4+x}\iff 2-\sqrt{3+x}\lt 4+x\iff \sqrt{3+x}\gt -2-x.$$ For the third, we have $$\sqrt{2x^2-7x+6}\gt 0\ \ \text{and}\ \ \sqrt{2x^2-7x+6}\not=1\ \ \text{and}\ \ \frac x3\gt 0$$ and we have $$\log_{\sqrt{2x^2-7x+6}}{\left(\frac x3\right)}\gt 0\iff \log_{\sqrt{2x^2-7x+6}}{\left(\frac x3\right)}\gt \log_{\sqrt{2x^2-7x+6}}1.$$ Now, separate it into two cases as $$0\lt \sqrt{2x^2-7x+6}\lt 1\ \ \text{or}\ \ \sqrt{2x^2-7x+6}\gt 1.$$

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