Cho biểu thức: 

- Hướng dẫn giải

1. Điều kiện : $0\le x\ne 1$

Ta có:

$\begin{array}{l}P=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}\sqrt{a}\right)\\ =\left(a+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\\ =\left(1+2\sqrt{a}+a\right)\left(1-2\sqrt{a}+a\right)\\ ={\left(1+\sqrt{a}\right)}^2{\left(1-\sqrt{a}\right)}^2\\ ={\left(1-a\right)}^2\end{array}$

2. Ta có:

$\begin{array}{l}P<7-4\sqrt{3}\\ \iff {\left(1-a\right)}^2<\mathrm{4.2.2}\sqrt{3}+{\left(\sqrt{3}\right)}^2\\ \iff {\left(1-a\right)}^2<{\left(2-\sqrt{3}\right)}^2\\ \iff -2+\sqrt{3}<1-a<2-\sqrt{3}\\ \iff -3+\sqrt{3}<-a<1-\sqrt{3}\\ \iff \sqrt{3}-1<a<3-\sqrt{3}\end{array}$

Vậy với $\sqrt{3}-1<a<3-\sqrt{3}$ thì $P<7-4\sqrt{3}$