2重根号のはずし方
2重根号とは
根号の中に根号があるもの
2重根号 \(\sqrt {p+2\sqrt {q}}\) のはずし方
\(=5+2\sqrt {6}\)
\(=\sqrt {(\sqrt {3}+\sqrt {2})^2}\)
\(=\sqrt {3}+\sqrt {2}\)
\((\sqrt {a}+\sqrt {b})^2=(a+b)+2\sqrt {ab}\)
和が \(5\) 、積が \(6\) であるから、
\(a=3\) , \(b=2\) よって、
\((\sqrt {a}-\sqrt {b})^2=\sqrt {(a+b)-2\sqrt {ab}}\)
まとめ
\(\sqrt {(a+b)+2\sqrt {ab}}=\sqrt {(\sqrt {a}+\sqrt {b})^2}\)
\(=\sqrt {a}+\sqrt {b}\)
\(\sqrt {(a+b)-2\sqrt {ab}}=\sqrt {(\sqrt {a}-\sqrt {b})^2}\)
\(=\sqrt {a}-\sqrt {b}\)
👉 \(a>b>0\) のとき、\(\sqrt {a}-\sqrt {b}>0\)
練習問題
練習問題 次の式を簡単にせよ。
⑴ \(\sqrt {6+2\sqrt {5}}\)
⑵ \(\sqrt {7-2\sqrt {12}}\)
⑶ \(\sqrt {12+\sqrt {44}}\)
⑷ \(\sqrt {14-6\sqrt {5}}\)
⑸ \(\sqrt {2+\sqrt {3}}\)
解答
⑴ \(\sqrt {6+2\sqrt {5}}\)
\(=\sqrt {(\sqrt {5}+\sqrt {1})^2}\)
\(=\sqrt {5}+\sqrt {1}\)
\(=\sqrt {5}+1\)
⑵ \(\sqrt {7-2\sqrt {12}}\)
\(=\sqrt {(\sqrt {4}-\sqrt {3})^2}\)
\(=\sqrt {4}-\sqrt {3}\)
\(=2-\sqrt {3}\)
⑶ \(\sqrt {12+\sqrt {44}}\)
\(=\sqrt {12+\sqrt {2^2\cdot 11}}\)
\(=\sqrt {12+2\sqrt {11}}\)
\(=\sqrt {(\sqrt {11}+\sqrt {1})^2}\)
\(=\sqrt {11}+\sqrt {1}\)
\(=\sqrt {11}+1\)
⑷ \(\sqrt {14-6\sqrt {5}}\)
\(=\sqrt {14-2\cdot 3\sqrt {5}}\)
\(=\sqrt {14-2\sqrt {3^2\cdot 5}}\)
\(=\sqrt {14-2\sqrt {45}}\)
\(=\sqrt {(\sqrt {9}-\sqrt {5})^2}\)
\(=\sqrt {9}-\sqrt {5}\)
\(=3-\sqrt {5}\)
⑸ \(\sqrt {2+\sqrt {3}}\)
\(=\)\(\sqrt {\frac {4+2\sqrt {3}}{2}}\)
\(=\)\(\frac {\sqrt {4+2\sqrt {3}}}{\sqrt {2}}\)
\(=\)\(\frac {\sqrt {3}+1}{\sqrt {2}}\)
\(=\)\(\frac {\sqrt {6}+\sqrt {2}}{2}\)