Question

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Answer 1

(a) Với biểu thức \(A:\)

\(A=\left(2+\sqrt{3}\right)\sqrt{2}-\sqrt{8}-\dfrac{\sqrt{666}}{\sqrt{111}}+\dfrac{1}{2}\)

\(=2\sqrt{2}+\sqrt{6}-2\sqrt{2}-\sqrt{\dfrac{666}{111}}+\dfrac{1}{2}\)

\(=\sqrt{6}-\sqrt{6}+\dfrac{1}{2}=\dfrac{1}{2}\)

Với biểu thức \(B\left(x\ge0;x\ne4\right):\)

\(B=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)

\(=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

\(=\dfrac{2-\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

\(=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

\(=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{\sqrt{x}+2}.\)

Vậy: \(A=\dfrac{1}{2};B=\dfrac{2}{\sqrt{x}+2}\left(x\ge0;x\ne4\right).\)

 

(b) Xét \(x=\sqrt{24-8\sqrt{5}}\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\left(TM\right)\).

Ta có: \(x=\sqrt{24-8\sqrt{5}}=\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot2+2^2}\)

\(=\sqrt{\left(2\sqrt{5}-2\right)^2}=2\sqrt{5}-2\)

Thay vào \(B\), ta được:

\(B=\dfrac{2}{\sqrt{x}+2}=\dfrac{2}{\sqrt{2\sqrt{5}-2}+2}\)

\(=\dfrac{2\left(\sqrt{2\sqrt{5}-2}-2\right)}{2\sqrt{5}-2-2^2}=\dfrac{2\left(\sqrt{2\sqrt{5}-2}-2\right)}{2\sqrt{5}-6}\)

\(=\dfrac{\sqrt{2\sqrt{5}-2}-2}{\sqrt{5}-3}.\)