Giải thích các bước giải:

1.Ta có:

$\begin{aligned}
& x = 0,49 \implies \sqrt{x} = \sqrt{0,49} = 0,7 \\
& M = \dfrac{\sqrt{x}-1}{\sqrt{x}} \\
& M = \dfrac{0,7 - 1}{0,7} \\
& M = \dfrac{-0,3}{0,7} \\
& M = -\dfrac{3}{7}
\end{aligned}$

2.Ta có:

$\begin{aligned}
P &= \dfrac{\sqrt{x}-2}{\sqrt{x}+1} + \dfrac{2+8\sqrt{x}}{x-1} - \dfrac{2}{1-\sqrt{x}} \\
P &= \dfrac{\sqrt{x}-2}{\sqrt{x}+1} + \dfrac{2+8\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)} + \dfrac{2}{\sqrt{x}-1} \\
P &= \dfrac{(\sqrt{x}-2)(\sqrt{x}-1) + (2+8\sqrt{x}) + 2(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)} \\
P &= \dfrac{(x - \sqrt{x} - 2\sqrt{x} + 2) + 2 + 8\sqrt{x} + 2\sqrt{x} + 2}{(\sqrt{x}-1)(\sqrt{x}+1)} \\
P &= \dfrac{x - 3\sqrt{x} + 2 + 2 + 8\sqrt{x} + 2\sqrt{x} + 2}{(\sqrt{x}-1)(\sqrt{x}+1)} \\
P &= \dfrac{x + 7\sqrt{x} + 6}{(\sqrt{x}-1)(\sqrt{x}+1)} \\
P &= \dfrac{x + \sqrt{x} + 6\sqrt{x} + 6}{(\sqrt{x}-1)(\sqrt{x}+1)} \\
P &= \dfrac{\sqrt{x}(\sqrt{x}+1) + 6(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)} \\
P &= \dfrac{(\sqrt{x}+1)(\sqrt{x}+6)}{(\sqrt{x}-1)(\sqrt{x}+1)} \\
P &= \dfrac{\sqrt{x}+6}{\sqrt{x}-1} \quad (\text{đpcm})
\end{aligned}$

3.Ta có:

$\begin{aligned}
& Q = M \cdot P + \dfrac{x-5}{\sqrt{x}} \\
& Q = \left( \dfrac{\sqrt{x}-1}{\sqrt{x}} \right) \cdot \left( \dfrac{\sqrt{x}+6}{\sqrt{x}-1} \right) + \dfrac{x-5}{\sqrt{x}} \\
& Q = \dfrac{\sqrt{x}+6}{\sqrt{x}} + \dfrac{x-5}{\sqrt{x}} \\
& Q = \dfrac{\sqrt{x} + 6 + x - 5}{\sqrt{x}} \\
& Q = \dfrac{x + \sqrt{x} + 1}{\sqrt{x}} \\
\\
& \text{Xét hiệu } Q - 3: \\
& Q - 3 = \dfrac{x + \sqrt{x} + 1}{\sqrt{x}} - 3 \\
& Q - 3 = \dfrac{x + \sqrt{x} + 1 - 3\sqrt{x}}{\sqrt{x}} \\
& Q - 3 = \dfrac{x - 2\sqrt{x} + 1}{\sqrt{x}} \\
& Q - 3 = \dfrac{(\sqrt{x}-1)^2}{\sqrt{x}} \\
\\
& \text{Với } x > 0, x \neq 1: \\
& \begin{cases} (\sqrt{x}-1)^2 > 0 \\ \sqrt{x} > 0 \end{cases} \\
\implies & \dfrac{(\sqrt{x}-1)^2}{\sqrt{x}} > 0 \\
\implies & Q - 3 > 0 \\
\implies & Q > 3
\end{aligned}$