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a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
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 : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
a) \(=\left(\frac{x+2\sqrt{x}-7}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{1-\sqrt{x}}{\sqrt{x}-3}\right)\div\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{x+2\sqrt{x}-7}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{x+2\sqrt{x}-7}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{3-2\sqrt{x}-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{x+2\sqrt{x}-7+3-2\sqrt{x}-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-3}\)
b) Để \(P\left(\sqrt{x}-3\right)=\left|x-3\right|\)
=> \(\frac{\sqrt{x}-1}{\sqrt{x}-3}\cdot\left(\sqrt{x}-3\right)=\left|x-3\right|\)(\(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\))
=> \(\sqrt{x}-1=\left|x-3\right|\)
=> \(\orbr{\begin{cases}\sqrt{x}-1=x-3\left(x\ge3\right)\\\sqrt{x}-1=3-x\left(1\le x< 3\right)\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=\frac{9-\sqrt{17}}{2}\end{cases}}\)
c) Em chịu T.T
\(đkxđ\Leftrightarrow x\ge0\)
\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}}:\frac{x-1-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(x-1\right)\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}.\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-1}{\sqrt{x}}\)
\(b,P.\sqrt{x}=6\sqrt{x}-3-\sqrt{x}-4\)
\(\Rightarrow\frac{x-1}{\sqrt{x}}.\sqrt{x}=5\sqrt{x}-7\)
\(\Rightarrow x-5\sqrt{x}+6=0\)
\(\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=3\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=9\end{cases}}}\)
Vậy \(x\in\left\{4;9\right\}\)