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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 : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{-\left(2x+\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x\sqrt{x}+x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\dfrac{-2x^2+x\sqrt{x}-2\sqrt{x}+1+2x^2-x\sqrt{x}-2x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{-2x-\sqrt{x}+1}\)
\(=\dfrac{-\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{-\sqrt{x}\left(2x+\sqrt{x}-1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
b: Thay \(x=17-12\sqrt{2}=\left(3-2\sqrt{2}\right)^2\) vào A, ta được:
\(A=\dfrac{17-12\sqrt{2}-\sqrt{2}+1+1}{3-2\sqrt{2}}=\dfrac{19-13\sqrt{2}}{3-2\sqrt{2}}=5-\sqrt{2}\)
a, \(P=\left(\frac{\sqrt{x}}{x\sqrt{x}-1}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b, Vì x > 1, g/s : Thay x = 4 vào P ta được :
\(\frac{\sqrt{4}+1}{\sqrt{4}-1}=\frac{3}{1}=3\)
Thay x = 4 vào căn P ta được : \(\sqrt{\frac{\sqrt{4}+1}{\sqrt{4}-1}}=\sqrt{3}\)
mà \(3>\sqrt{3}\Rightarrow P>\sqrt{P}\)với x > 1
\(A=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left[\left(\frac{1}{2\sqrt{x}}\right)^2-2.\frac{1}{2\sqrt{x}}.\frac{\sqrt{x}}{2}+\left(\frac{\sqrt{x}}{2}\right)^2\right]\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right]\left(\frac{1}{4x}-\frac{1}{2}+\frac{x}{4}\right)\)
\(\Leftrightarrow A=\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\left(\frac{1}{4x}-\frac{2x}{4x}+\frac{x^2}{4x}\right)\)
\(\Leftrightarrow A=\frac{-4\sqrt{x}}{x-1}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{4\sqrt{x}}{1-x}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{1-x}{\sqrt{x}}\)
b) \(\frac{A}{\sqrt{x}}>1\)
\(\Leftrightarrow\frac{1-x}{\frac{\sqrt{x}}{\sqrt{x}}}>1\)
\(\Leftrightarrow1-x>1\Leftrightarrow x< 0\)
ĐK: \(x\ge0;x\ne1\)
a) \(P=\frac{\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(P=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}.\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
Để \(P=\sqrt{x}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}\Leftrightarrow\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)\)\(\sqrt{x}+1\Leftrightarrow x-\sqrt{x}\Leftrightarrow-x+2\sqrt{x}+1=0\)
\(\Leftrightarrow-\left(x-2\sqrt{x}+1\right)+2=0\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=\sqrt{2}\\\sqrt{x}-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{2}+1\\\sqrt{x}=-\sqrt{2}+1\end{cases}\Leftrightarrow}x=3\pm2\sqrt{2}}\)
b) Với \(x>1\)thì \(P>0\)
Ta dễ thấy \(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}>1\)
Ta có: \(P>0;P>1\)\(\Rightarrow P\left(P-1\right)>0\Leftrightarrow P^2>P\Leftrightarrow P>\sqrt{P}\)