多重二次根式的化简
<h1 id="复习单层二次根式的化简">复习：单层二次根式的化简</h1><blockquote>
<p>化简：</p>
<p></p><div class="math display">\[\sqrt {54188a^{114514}b^{45}}
\]</div><p></p></blockquote>
<p>这种题目已经是基本操作了。解析如下：</p>
<blockquote>
<p></p><div class="math display">\[解：原式=\sqrt {2^2\cdot 13547\cdot (a^{57257})^2\cdot (b^{24})^2\cdot b}
\]</div><p></p><p></p><div class="math display">\[=2a^{57257}b^{24}\sqrt {13547b}
\]</div><p></p></blockquote>
<p>任意的单层二次根式，都可以通过这种方式化简而来。</p>
<h1 id="多重二次根式">多重二次根式</h1>
<h2 id="引入">引入</h2>
<blockquote>
<p>化简：</p>
<p></p><div class="math display">\[\sqrt {4-2\sqrt 3}
\]</div><p></p></blockquote>
<p>这是什么根式套根式的玩意儿啊！怎么化简呢？</p>
<p>我们注意到，要想化简一个双层根式，至少应该先把一层根号去掉。</p>
<p>那你觉得<span class="math inline">\(\sqrt 3\)</span>更好去根号，还是<span class="math inline">\(\sqrt {4-2\sqrt 3}\)</span>更好去根号呢？<br>
当然是希望去掉外层根号。</p>
<p>那么，在我们现在已学的公式中，有什么东西能把一个根号去掉呢？</p>
<p></p><div class="math display">\[\sqrt {a^2}=|a|
\]</div><p></p><p></p><div class="math display">\[(\sqrt a)^2=a\{a>0\}
\]</div><p></p><p>请从这两个公式上思考一下。<br>
显然，把<strong>内部的<span class="math inline">\(4-2\sqrt 3\)</span>化成某个根式的平方即可</strong>。完全平方公式是最好的选择。</p>
<p>恰好<span class="math inline">\(2\sqrt 3\)</span>可以看做公式中的<span class="math inline">\(2ab\)</span>。于是，我们希望找到两个式子<span class="math inline">\(a\)</span>、<span class="math inline">\(b\)</span>，使得<span class="math inline">\(ab=\sqrt 3\)</span>，且要满足<span class="math inline">\(a^2+b^2=4\)</span>。</p>
<p>你看着，<span class="math inline">\(ab\)</span>都出现根号了，那<span class="math inline">\(a\)</span>、<span class="math inline">\(b\)</span>能不是根式吗？<br>
因此，目标转化为：</p>
<blockquote>
<p>求两个整数<span class="math inline">\(a\)</span>、<span class="math inline">\(b\)</span>，使得<span class="math inline">\(\sqrt a\cdot \sqrt b=\sqrt {ab}=\sqrt 3\)</span>即<span class="math inline">\(ab=3\)</span>，且<span class="math inline">\((\sqrt a)^2+(\sqrt b)^2=a+b=4\)</span>。</p>
</blockquote>
<p>挨个试一试也出来了，<span class="math inline">\(a=1\)</span>，<span class="math inline">\(b=3\)</span>。最终，原式</p>
<p></p><div class="math display">\[=\sqrt {(\sqrt 1)^2-2\sqrt {1\cdot 3}+(\sqrt 3)^2}
\]</div><p></p><p></p><div class="math display">\[=\sqrt {(\sqrt 1+\sqrt 3)^2}
\]</div><p></p><p></p><div class="math display">\[=1+\sqrt 3
\]</div><p></p><p>好神奇呀。</p>
<h2 id="化简方法完全平方公式">化简方法：完全平方公式</h2>
<p>如引入中所示。<br>
下面拓展一些变式。</p>
<blockquote>
<p>出题人看到了一个题目。</p>
<p>化简：</p>
<p></p><div class="math display">\[\sqrt {8-2\sqrt {12}}
\]</div><p></p></blockquote>
<p>你会做吗？当然会做啊，原式</p>
<p></p><div class="math display">\[=\sqrt {2+2\sqrt{2\cdot 6}+6}
\]</div><p></p><p></p><div class="math display">\[=\sqrt {(\sqrt 2+\sqrt 6)^2}
\]</div><p></p><p></p><div class="math display">\[=\sqrt 2+\sqrt 6
\]</div><p></p><blockquote>
<p>出题人：太简单了太简单了，不符合我<s>哈基米</s>的气质。我来改一改。</p>
<p>化简：</p>
<p></p><div class="math display">\[\sqrt {8-4\sqrt 3}
\]</div><p></p></blockquote>
<p>刚刚会做，难道现在不会做了吗？原式</p>
<p></p><div class="math display">\[=\sqrt {8-2\sqrt {12}}=\sqrt 2+\sqrt 6
\]</div><p></p><p><strong>变式<span class="math inline">\(1\)</span>：把后面的根式化到最简，故意让你看不出来。</strong><br>
<strong>对策<span class="math inline">\(1\)</span>：只要看见<span class="math inline">\(a+b\sqrt c\)</span>的形式，尝试把<span class="math inline">\(b\)</span>化成<span class="math inline">\(2\)</span>。</strong></p>
<blockquote>
<p>出题人：可恶，这次得分率还是那么高。我再来改改。</p>
<p>化简：</p>
<p></p><div class="math display">\[\sqrt {2-\sqrt 3}
\]</div><p></p></blockquote>
<p>这下彻底炸了，<span class="math inline">\(b\)</span>直接消失，还怎么化成<span class="math inline">\(2\)</span>？</p>
<p>你忘了。<span class="math inline">\(b\)</span>其实是<span class="math inline">\(1\)</span>，被隐藏了。原式</p>
<p></p><div class="math display">\[=\sqrt {\frac {4-2\sqrt 3} 2}
\]</div><p></p><p></p><div class="math display">\[=\sqrt {\frac {(1-\sqrt 3)^2} 2}
\]</div><p></p><p></p><div class="math display">\[=\frac {1-\sqrt 3} {\sqrt 2}
\]</div><p></p><p>最后再来一个分母有理化。</p>
<p></p><div class="math display">\[=\frac {\sqrt 2-\sqrt 6} 2
\]</div><p></p><p><strong>变式<span class="math inline">\(2\)</span>：把外层根式除个<span class="math inline">\(d\)</span>，正好把<span class="math inline">\(b\)</span>隐藏。</strong><br>
<strong>对策<span class="math inline">\(2\)</span>：看不见<span class="math inline">\(b\)</span>，立刻外层乘个<span class="math inline">\(2\)</span>，分子常规化，分母有理化。</strong></p>
<h1 id="题目汇编">题目汇编</h1>
<h2 id="1-又是数形结合">$1. $【又是数形结合】</h2>
<blockquote>
<p>已知非负实数<span class="math inline">\(a\)</span>、<span class="math inline">\(b\)</span>、<span class="math inline">\(c\)</span>满足<span class="math inline">\(a+b+c=8\)</span>。求<span class="math inline">\(\sqrt {a^2+1}+\sqrt {b^2+4}+\sqrt {c^2+9}\)</span>的最小值。</p>
</blockquote>
<p>数形结合万事休。<br>
<img src="https://cdn.luogu.com.cn/upload/image_hosting/zgfxkbz9.png" alt="" loading="lazy"></p>
<p></p><div class="math display">\[\sqrt {a^2+1}+\sqrt {b^2+4}+\sqrt {c^2+9}\geq \sqrt {6^2+8^2}=10
\]</div><p></p><br><br>
来源：https://www.cnblogs.com/ZzqMath-Coding/p/18895244
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