1) Rút gọn a) `frac{sqrt{6}+sqrt{14}}{2sqrt{3}+sqrt{28}}` b) `frac{sqrt{2}+sqrt{3}+sqrt{6}+sqrt{8}+sqrt{16}}{sqrt{2}+sqrt{3}+sqrt{4}}` c)`frac{x+1}{x-

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1. Đáp án:

   a/ $\dfrac{\sqrt{2}}{2}$

   b/ $1+\sqrt{2}$

   c/ $\sqrt{\dfrac{(x+1)(x-2)}{x-1}}$

   d/ $\sqrt{x-1}$

   Giải thích các bước giải:

   a/ $\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}$

   $=\dfrac{\sqrt{6}+\sqrt{14}}{\sqrt{12}+\sqrt{28}}$

   $=\dfrac{\sqrt{6}+\sqrt{14}}{\sqrt{2}(\sqrt{6}+\sqrt{14})}$

   $=\dfrac{1}{\sqrt{2}}$

   $=\dfrac{\sqrt{2}}{2}$

   b/ $\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

   $=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

   $=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

   $=1+\dfrac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

   $=1+\sqrt{2}$

   c/ $\dfrac{x+1}{x-1}.\sqrt{\dfrac{x^2-3x+2}{x+1}}$ $(2)$

   $\text{ĐKXĐ: $\dfrac{x^2-3x+2}{x+1} \geq 0$ (1) và $x \neq ±1$}$

   $(1) ⇔ \left[ \begin{array}{l}\begin{cases}(x-2)(x-1)\geq 0 \\x+1 \geq 0\end{cases}\\\begin{cases}(x-2)(x-1) \leq 0 \\x+1 \leq 0\end{cases}\end{array} \right.$

   $⇔ \left[ \begin{array}{l}\begin{cases}\left[ \begin{array}{l}x \leq 1\\x\geq 2\end{array} \right.\\x\geq -1\end{cases}\\\begin{cases}\left[ \begin{array}{l}1 \leq x \leq 2\\x\leq-1\end{array}\right.\\x\leq -1\end{cases}\end{array} \right.$

   $⇔ \left[ \begin{array}{l}\left[ \begin{array}{l}-1 \leq x \leq 1\\x \geq 2\end{array} \right.\\x \leq -1\end{array} \right.$

   $⇔ \left[ \begin{array}{l}-1 < x < 1\\x \geq 2\\ x <-1\end{array} \right.$ $\text{(do $x \neq ±1$)}$

   $\text{⇒ Bthức xác định với mọi $x \neq -1$ và k nằm trong khoảng $1 \leq x<2$}$

   $(2)=\dfrac{x+1}{x-1}.\sqrt{\dfrac{x^2-2x-x+2}{x+1}}$

   $=\dfrac{x+1}{x-1}.\sqrt{\dfrac{(x-2)(x-1)}{x+1}}$

   $=\dfrac{(\sqrt{x+1})^2}{(\sqrt{x-1})^2}.\dfrac{\sqrt{x-2}.\sqrt{x-1}}{\sqrt{x+1}}$

   $=\dfrac{\sqrt{x+1}.\sqrt{x-2}}{\sqrt{x-1}}$

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

   d/ $\sqrt{(x-1)(4x^2-4x^2+1)}$

   $\text{ĐKXĐ: $x \geq 1$}$

   $=\sqrt{x-1}$

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