Bài 1:
1) \(A=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{3-1}=2+\sqrt{3}-\sqrt{3}-1=1\)
2) ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}-3\sqrt{x-2}=-1\Leftrightarrow-2\sqrt{x-2}=-1\Leftrightarrow x-2=\dfrac{1}{4}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\)
Bài 2:
a) \(A=\dfrac{2\sqrt{25}-1}{\sqrt{25}-1}=\dfrac{2.5-1}{5-1}=\dfrac{9}{4}\)
b) \(B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
c) \(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}=2-\dfrac{3}{\sqrt{x}+1}< 1\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+1}>1\Leftrightarrow0< \sqrt{x}+1< 3\Leftrightarrow0\le x\le4,x\ne1\)
Bài 1:
\(1,A=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{2}=\sqrt{3}+2-\sqrt{3}-1=1\\ 2,ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}-3\sqrt{x-2}=-1\\ \Leftrightarrow-2\sqrt{x-2}=-1\\ \Leftrightarrow\sqrt{x-2}=\dfrac{1}{2}\\ \Leftrightarrow x-2=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\)
Bài 2:
\(a,A=\dfrac{10-1}{5-1}=\dfrac{9}{4}\\ b,B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ c,P=AB=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\\ \Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}-1< 0\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\left(\sqrt{x}+1>0\right)\\ \Leftrightarrow0\le x< 4\\ \Leftrightarrow x\in\left\{0;1;2;3\right\}\)
