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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...........
\(\text{ĐKXĐ: }x\ge0;x\ne1\)
\(A=\frac{x-2\sqrt{x}}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\frac{1-2x-2\sqrt{x}}{x^2-\sqrt{x}}\)
\(=\frac{x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\frac{1-2x-2\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{1-2x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-3x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
ủa đề có sai ko zạy
\(=\frac{\sqrt{x}.\left(x-3\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)