Giải thích các bước giải:

a.ĐKXĐ: $x\ge 0, x\ne 1$

Ta có:

$B7=(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+1})\cdot \dfrac{x^2-2x+1}2$

$\to B7=(\dfrac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{\sqrt{x}+2}{(\sqrt{x}-1)^2})\cdot \dfrac{(x-1)^2}2$

$\to B7=(\dfrac{(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}-1)^2(\sqrt{x}+1)}-\dfrac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}+1)(\sqrt{x}-1)^2})\cdot \dfrac{(x-1)^2}2$

$\to B7=\dfrac{(\sqrt{x}-2)(\sqrt{x}-1)-(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}+1)(\sqrt{x}-1)^2}\cdot \dfrac{(x-1)^2}2$

$\to B7=\dfrac{ -6\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)^2}\cdot \dfrac{(\sqrt{x}-1)^2(\sqrt{x}+1)^2}2$

$\to B7=-3\sqrt{x}(\sqrt{x}+1)$

b.Vì $0<x<1\to \sqrt{x}, \sqrt{x}+1>0$

$\to -3\sqrt{x}(\sqrt{x}+1)<0$

$\to B7<0$

c.Khi $x=0.16\to \sqrt{x}=0.4$

$\to B7=-3\cdot 0.4(0.4+1)$

 $\to B7=-1.68$