@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm

còn câu b kl sai r nha

a/ \(Q=\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right].\frac{2}{\sqrt{x}-1}\)

       \(=\frac{x+2-x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\left(\sqrt{x}-1\right)}\)

       \(=\frac{\left(x-2\sqrt{x}+1\right).2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\) 

       \(=\frac{2}{x+\sqrt{x}+1}\)

b/ Ta có: \(x+\sqrt{x}+1=x+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

                      \(\Rightarrow Q=\frac{2}{x+\sqrt{x}+1}>0\). 

                                                                                  Vậy Q > 0

a) A= \(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{2}\right)\) (x ≥ 0; x ≠ 4)

= \(\left(\frac{x+2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{2}\)

=\(\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

=\(\left(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

= \(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\cdot\frac{2}{\sqrt{x}-1}\)

=\(\frac{2}{x+\sqrt{x}+1}\)

b) Ta có: x ≥ 0 ⇒ \(\sqrt{x}\) ≥ 0

⇒x+\(\sqrt{x}\)+1 ≥ 1 > 0

mà 2 > 0

⇒ A > 0 (1)

Ta có:

\(x+\sqrt{x}+1\) ≥ 1

⇒ \(\frac{1}{x+\sqrt{x}+1}\) ≤ 1

⇒\(\frac{2}{x+\sqrt{x}+1}\) ≤ 2

⇒A ≤ 2 (2)

Từ (1) và (2) => 0 < A ≤ 2

\( a)A = \dfrac{{a - \sqrt a - 6}}{{4 - a}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{a + 2\sqrt a - 3\sqrt a - 6}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 3}}{{\sqrt a - 2}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 2}}{{\sqrt a - 2}} = - 1 \)

\( b)B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{x - 1}}\\ B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{\sqrt x + 1 + \sqrt x - 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{2}{{\sqrt x + 1}} \)