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\(a,\)\(T=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x}^3-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\)\(\frac{\sqrt{x}^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)\(-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
1. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}-2+\sqrt{3}=VP\)
Bài 1.
Ta có : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
Ta có: \(P=\frac{\sqrt{x}-4}{\sqrt{x}}\times\frac{x+\sqrt{x}+1}{\sqrt{x}-4}\)
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)\(\left(ĐK:x>0\right)\)
Ta lấy \(P-2=\frac{x+\sqrt{x}+1}{\sqrt{x}}-2\)
\(=\frac{x+\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{3}{4}}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{x}}\)
Vì \(x>0\Rightarrow\sqrt{x}>0\)
\(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{x}}>0\)
\(\Rightarrow P-2>0\)
\(\Rightarrow P>2\)
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