Ta có: a√a = √(a².a) = (√a)³ 
=> 1 - a√a = 1 - (√a)³ = (1 - √a)(a + √a + 1) (1) 
Tương tự: 1 + a√a = 1 + (√a)³ = (1 + √a)(a - √a + 1) (2) 
Từ (1) và (2) => [ (1-a√a/1-√a+√a).(1+a√a/1+√a-√a) + 1 ]. 
= [(1 - √a)(a + √a + 1)/(1 - √a) + √a].[(1 + √a)(a - √a + 1)/(1 + √a) - √a ] +1 
=(a + √a + 1 + √a)(a - √a + 1- √a) + 1 
= (a + 2√a + 1)(a - 2√a + 1) + 1 
= (√a + 1)²(√a - 1)² +1 
= [(√a + 1)(√a - 1)]² + 1 
= (a - 1)² + 1 
= a² - 2a + 1 + 1 
= a² - 2a + 2 
=> [ (1-a√a/1-√a+√a).(1+a√a/1+√a-√a) + 1 ] = a² - 2a + 2 (3) 
Áp dụng (3) vào A ta được A = [(1 - a)²]/(a² - 2a + 2) 
<=> A = (a² - 2a + 1)/(a² - 2a + 2) 

\(ĐKXĐ:a\ge0\)

\(A=\left(\frac{2\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}+1}+\frac{1}{\sqrt{a}+1}\right):\left(1+\frac{\sqrt{a}}{a+1}\right)\)

\(\Leftrightarrow A=\left(\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}+\frac{1}{\sqrt{a}+1}\right):\frac{a+\sqrt{a}+1}{a+1}\)

\(\Leftrightarrow A=\frac{2\sqrt{a}+a+1}{\left(a+1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a+1}{a+\sqrt{a}+1}\)

\(\Leftrightarrow A=\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(a+\sqrt{a}+1\right)}\)

\(\Leftrightarrow A=\frac{\sqrt{a}+1}{a+\sqrt{a}+1}\)

Câu C : Lần đầu làm dạng này :))

Xét hiệu A - 2 , ta có :

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

Ta thấy :

+) Do \(a\ge0\)\(\Rightarrow2\sqrt{a}\left(1-2\sqrt{a}\right)\le0\)

+) a khác 1 ; \(a\ge0\)=> 2a + 1 > 0

\(\Rightarrow\frac{2\sqrt{a}\left(1-2\sqrt{a}\right)}{2a+1}\le0\)

\(\Leftrightarrow A< 2\)

P/s : sai bỏ qua :))

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

ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)

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

\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{a+2\sqrt{a}+1+a-\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(A=\frac{2}{\sqrt{a}-1}\div\frac{2a+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

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

\(A=\frac{2\left(\sqrt{a}+1\right)}{2a+1}\)

b) \(a=1-\frac{\sqrt{3}}{2}=\frac{2}{2}-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}\)( tmđk )

Rồi từ đây thế vô :)

c) Nhờ cao nhân làm tiếp chứ em mới lớp 8 thôi ạ :(

ĐKXĐ: Bạn tự làm nha 

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

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

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

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

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

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

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

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

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

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

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

\(=\frac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)

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

= \(\left[\left(\frac{\sqrt{a}}{2}\right)^2-2\frac{\sqrt{a}}{2}\frac{1}{2\sqrt{a}}+\left(\frac{1}{2\sqrt{a}}\right)^2\right]\).\(\left[\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{a-1}\right]\)

=\(\left(\frac{a}{4}-\frac{1}{2}+\frac{1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1\right)^2}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)^2}{a-1}\right]\)

=\(\left(\frac{a^2}{4a}-\frac{2a}{4a}+\frac{1}{4a}\right)\).\(\left[\frac{\left[\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)\right]\cdot\left[\left(\sqrt{a}-1\right)+\left(\sqrt{a}+1\right)\right]}{a-1}\right]\)

=\(\left(\frac{a^2-2a+1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1-\sqrt{a}+1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right]\)

=\(\frac{\left(a-1\right)^2}{1}\).\(\frac{-4\sqrt{a}}{a-1}\)

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

hok tốt 

\(A=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]\div\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\)

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

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

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