Câu 2: 

Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)

\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)

\(=1-a\)

Câu 1: 

Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)

\(=1\)

P = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-\sqrt{a}}\)

= \(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)  

= 1

\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)

\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)

\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)

\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)

\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)

Bài 1: 

a) Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)

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

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

b) Để Q dương thì \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)

mà \(3\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ

nên \(\sqrt{a}-2>0\)

\(\Leftrightarrow\sqrt{a}>2\)

hay a>4

Kết hợp ĐKXĐ,ta được: a>4

Vậy: Để Q dương thì a>4

\(a,\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)

\(b,A=\dfrac{\sqrt{a}}{\sqrt{a}-5}-\dfrac{10\sqrt{a}}{a-25}-\dfrac{5}{\sqrt{a}+5}\)

\(\Rightarrow A=\dfrac{\sqrt{a}\left(\sqrt{a}+5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\left(\sqrt{a}-5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a+5\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\sqrt{a}-25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a-10\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{\left(\sqrt{a}-5\right)^2}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)

a: \(=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)

b: \(A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}=\dfrac{\left(\sqrt{a}-5\right)^2}{a-25}=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)

\(M=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right)=\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)=a-1\)

A=\(\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(a-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(a+\sqrt{a}\right)}{\left(a-1\right)}\right]\)::::::::\(\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)

=\(\left[\dfrac{1}{\sqrt{a}-1}\right]:\left(\dfrac{2\sqrt{a}}{a-1}\right)\)=\(\dfrac{\sqrt{a}-1}{2\sqrt{a}}\)

=\(\dfrac{a^2+a\sqrt{a}+11a+6}{2\sqrt{a}\left(\sqrt{a}+2\right)}\)

Ta có: \(A=\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)

\(=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)

\(=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)

\(A=\left(\dfrac{1-a\sqrt{a}}{1-a\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\left(dkxd:a\ge0,a\ne1\right)\)

\(=\left(1+\sqrt{a}\right).\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)

\(=\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1-\sqrt{a}\right)}{\left(1-a\right)^2}\)

\(=\dfrac{\left(1-a\right)\left(1-\sqrt{a}\right)}{\left(1-a\right)^2}\)

\(=\dfrac{1-\sqrt{a}}{1-a}\) 

Vậy \(A=\dfrac{1-\sqrt{a}}{1-a}\) với \(a\ge0,a\ne1\)

\(a.\sqrt{8}-2\sqrt{50}+\sqrt{18}=2\sqrt{2}-10\sqrt{2}+3\sqrt{2}=\sqrt{2}\left(2-10+3\right)=-5\sqrt{2}\)

\(b.\left(\dfrac{\sqrt{a}-a}{1-\sqrt{a}}+\sqrt{a}\right):\dfrac{2\sqrt{a}}{1+\sqrt{a}}\left(đk:a\ge0;a\ne1\right)\)

\(=\left(\sqrt{a}+\sqrt{a}\right).\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)

\(=2\sqrt{a}.\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)

\(=1+\sqrt{a}\)

(Chỗ điều kiện bài b mik thấy a = 0 cũng có thể là nghiệm nên mik sửa lại nhé)

b. \(=\left(\dfrac{\sqrt{a}-a+a\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)

\(=\left(\dfrac{2\sqrt{a}}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)

\(=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\)

\(=1-a\)