Giải giúp mình với ạ

Giải thích các bước giải:

a.Ta có:

$x=\dfrac14\to \sqrt{x}=\dfrac12$

$\to A=\dfrac{\dfrac12-1}{\dfrac12}=-1$

b.Ta có:

$B=\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac5{2-\sqrt{x}}-\dfrac{3\sqrt{x}+4}{x-2\sqrt{x}}$

$\to B=\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}-2)}+\dfrac5{\sqrt{x}-2}-\dfrac{3\sqrt{x}+4}{\sqrt{x}(\sqrt{x}-2)}$

$\to B=\dfrac{x-4}{\sqrt{x}(\sqrt{x}-2)}+\dfrac{5\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}-\dfrac{3\sqrt{x}+4}{\sqrt{x}(\sqrt{x}-2)}$

$\to B=\dfrac{x-4+5\sqrt{x}-3\sqrt{x}-4}{\sqrt{x}(\sqrt{x}-2)}$

$\to B=\dfrac{x+2\sqrt{x}-8}{\sqrt{x}(\sqrt{x}-2)}$

$\to B= \dfrac{(\sqrt{x}-2)(\sqrt{x}+4)}{\sqrt{x}(\sqrt{x}-2)}$

$\to B=\dfrac{\sqrt{x}+4}{\sqrt{x}}$

c.Ta có:

$P=\dfrac{\sqrt{x}-1}{\sqrt{x}}\cdot \dfrac{\sqrt{x}+4}{\sqrt{x}}$

$\to P=\dfrac{(\sqrt{x}-1)(\sqrt{x}+4)}{x}$

$\to P=\dfrac{x+3\sqrt{x}-4}{x}$

 Đặt $P=k, \sqrt{x}=t$

$\to k=\dfrac{t^2+3t-4}{t^2}$

$\to t^2+3t-4=kt^2$

$\to (k-1)t^2-3t+4=0$

$\to \Delta\ge 0$

$\to 3^2-4\cdot 4(k-1)\ge 0$

$\to k\le\dfrac{25}{16}$

$\to P\le \dfrac{25}{16}$

$\to GTLN_P=\dfrac{25}{16}$

$\to \dfrac{25}{16}=\dfrac{x+3\sqrt{x}-4}{x}$

$\to x=\dfrac{64}9$