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Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
\(B=\left(\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\right):\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
\(=\frac{2x+18\sqrt{x}-x-9\sqrt{x}}{x-9}\cdot\frac{\left(4-x\right)^2}{x\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+9\right)\left(4-x\right)^2}{x\sqrt{x}\left(x-9\right)}\)
\(=\frac{\left(\sqrt{x}+9\right)\left(4-x\right)^2}{\sqrt{x}\left(x-9\right)}\)
đkxđ: \(x\ge0;x\ne4\)
\(Q=\left[\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right]\div\left[\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)
\(Q=\left[\frac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\div\left[\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)
\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\div\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6\sqrt{x}}\)
\(Q=\frac{\left(x+9\right)\sqrt{x}}{6x}\)
\(Q=\frac{x\sqrt{x}+9\sqrt{x}}{6x}\)
đkxđ sửa tí thành \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
Điều kiện xác định \(0\le x\ne4\)
\(C=\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}+\frac{18\sqrt{x}}{4-x}\right):\frac{x+9}{4-x}\)
\(=\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-18\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{x+9}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+4\sqrt{x}+4-\left(x-4\sqrt{x}+4\right)-18\sqrt{x}}{-\left(x+9\right)}\)
\(=\frac{10\sqrt{x}}{x+9}\)