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1.
a. ĐKXĐ : x lớn hơn hoặc bằng 1/2
b. A\(\sqrt{2}\)= \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
= \(\sqrt{2x-1+1+2\sqrt{2x-1}}-\sqrt{2x-1+1-2\sqrt{2x-1}}\)
=\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
= \(\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)
Nếu \(x\ge1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(\sqrt{2x-1}-1\right)=2\)
\(\Rightarrow A=2\)
Nếu 1/2 \(\le x< 1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(1-\sqrt{2x-1}\right)=2\sqrt{2x-1}\)
Do đó : A= \(\sqrt{4x-2}\)
Vậy ............
2.
a. \(x\ge2\)hoặc x<0
b. A= \(2\sqrt{x^2-2x}\)
c. A<2 \(\Leftrightarrow\)\(2\sqrt{x^2-2x}< 2\Leftrightarrow\sqrt{x^2-2x}< 1\Leftrightarrow x^2-2x< 1\Leftrightarrow\left(x-1\right)^2< 2\)
\(-\sqrt{2}< x-1< \sqrt{2}\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\)
Kết hợp vs đk câu a , ta đc : \(1-\sqrt{2}< x< 0và2\le x< 1+\sqrt{2}\)
Vậy...........

a) A=\(\left(\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\right)\times\frac{x-4}{\sqrt{x}+3}\) (dkxd:x # 4 ,x# 9)
=> A=\(\frac{\sqrt{x}+2-\sqrt{x}+2}{x-4}\times\frac{x-4}{\sqrt{x}+3}\)
=> A=\(\frac{4}{\sqrt{x}+3}\)
b) A>1/2 <=> \(\frac{4}{\sqrt{x}+3}>\frac{1}{2}\Leftrightarrow\sqrt{x}+3< 8\Leftrightarrow\sqrt{x}< 5\Leftrightarrow x< 25\) (tmdkxd)
Vay .....

bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
Câu a.
\( A = 2x - 1 - \sqrt {{x^2} - x + \dfrac{1}{4}} \\ A = 2x - 1 - \sqrt {{{\left( {x - \dfrac{1}{2}} \right)}^2}} \\ A = 2x - 1 - \left( {x - \dfrac{1}{2}} \right)\\ A = 2x - 1 - x + \dfrac{1}{2}\\ A = x - \dfrac{1}{2} \)
Câu b.
\(A = \dfrac{3}{2} \Rightarrow x - \dfrac{1}{2} = \dfrac{3}{2} \Rightarrow x = \dfrac{3}{2} + \dfrac{1}{2} = 2 \)