30205中3・平方根・変形2【解】

平方根

変 形 2

計算問題

解 答 編

分母の有理化

$\;\;\begin{eqnarray}a>0,b>0\end{eqnarray}\;\;$のとき

$\;\;\begin{eqnarray}\sqrt{a^{2}b}&=&a\sqrt{b}\\[5pt]\sqrt{\frac{b}{a^{2}}}&=&\frac{\sqrt{b}}{a} \end{eqnarray}\;\;$

$\begin{eqnarray}(1)\;\;\sqrt{24}&=&\sqrt{4 \times 6}\\[3pt]&=&\sqrt{2^{2}\times 6}\\[3pt]&=&\sqrt{2^{2}}\times\sqrt{6}\\[3pt]&=&2\sqrt{6}\end{eqnarray}\;\;$

$\;\;2\sqrt{6}$


$\begin{eqnarray}(2)\;\;\sqrt{80}&=&\sqrt{16\times 5}\\[3pt]&=&\sqrt{4^{2}\times 5}\\[3pt]&=&\sqrt{4^{2}}\times\sqrt{5}\\[3pt]&=&4\sqrt{5}\end{eqnarray}\;\;$

$\;\;4\sqrt{5}$


$\begin{eqnarray}(3)\;\;\sqrt{200}&=&\sqrt{100\times 2}\\[3pt]&=&\sqrt{10^{2}\times 2}\\[3pt]&=&\sqrt{10^{2}}\times\sqrt{2}\\[3pt]&=&10\sqrt{2}\end{eqnarray}\;\;$

$\;\;10\sqrt{2}$


$\begin{eqnarray}(4)\;\;\sqrt{\frac{3}{49}}&=&\sqrt{\frac{3}{7^{2}}}\\[3pt]&=&\frac{\sqrt{3}}{\sqrt{7^{2}}}\\[3pt]&=&\frac{\sqrt{3}}{7}\end{eqnarray}\;\;$

$\;\;\dfrac{\sqrt{3}}{7}$


$\begin{eqnarray}(5)\;\;\sqrt{\frac{8}{169}}&=&\sqrt{\frac{2^{2}\times 2}{13^{2}}}\\[3pt]&=&\frac{\sqrt{2^{2}\times2}}{\sqrt{13^{2}}}\\[3pt]&=&\frac{2\sqrt{2}}{13}\end{eqnarray}\;\;$

$\;\;\dfrac{2\sqrt{2}}{13}$


$\begin{eqnarray}(6)\;\;\sqrt{0.002}&=&\sqrt{\frac{20}{10000}}\\[3pt]&=&\sqrt{\frac{2^{2}\times 5}{100^{2}}}\\[3pt]&=&\frac{\sqrt{2^{2}\times 5}}{\sqrt{100^{2}}}\\[3pt]&=&\frac{2\sqrt{5}}{100}\\[3pt]&=&\frac{\sqrt{5}}{50}\end{eqnarray}\;\;$

$\begin{eqnarray}《別解\,》\;\;\sqrt{0.002}&=&\sqrt{\frac{20}{10000}}\\[3pt]&=&\sqrt{\frac{5}{2500}}\\[3pt]&=&\sqrt{\frac{5}{50^{2}}}\\[3pt]&=&\frac{\sqrt{5}}{\sqrt{50^{2}}}\\[3pt]&=&\frac{\sqrt{5}}{50}\end{eqnarray}\;\;$

$\;\;\dfrac{\sqrt{5}}{50}$