合理化分母

合理化分母当我们从分数底部移动到顶部的任何分数力。 $\压裂{1}{\ sqrt {2}}$例如，它有一个无理数分母。我们很快就会看到它等于 $\压裂{\ sqrt {2}} {2}$．让我们看看具有非理性分母的分数。

考虑分数

$\压裂{一}{\ sqrt {b}}。$

将其写为完全有效

${对齐}\ \开始压裂{一}{\ sqrt {b}} & = \压裂{一}{\ sqrt {b}} \ * 1 \ \ & = \压裂{一}大概{b}}{\ \ * \压裂{\ sqrt {b}} {\ sqrt {b}} \ \ & = \压裂{\倍根号{b}} {b}。结束\{对齐}$

现在我们已经理性化了分母，大多数问题将依赖于这个简单的技术。

合理化的分母 $\ frac {1} {\ sqrt {2}}。$

---

运用我们所学的，我们拥有的

$\压裂{1}{\ sqrt{2}} = \压裂{1}{\ sqrt{2}} \ * \压裂{\ sqrt {2}} {\ sqrt{2}} = \压裂{\ sqrt{2}}{2}。\ _ \广场$

合理化的分母 $大概{\压裂{2}{\ \ sqrt {2}}}$．

---

我们有

${对齐}\ \开始dfrac {2} {\ sqrt {\ sqrt {2}}} & = \ dfrac{2×\√{\ sqrt {2}}} {\ sqrt {\ sqrt{2}}×\ sqrt {\ sqrt {2 }}}\\ & = \ dfrac{2 \√{\ sqrt{2}}}{\√6 {2 }}\\ & = \ dfrac{2 \√{\ sqrt{2}}×\ sqrt {2}} {\ sqrt{2}×\ sqrt {2 }}\\ & = \ dfrac{2 \√{2 \√{2}}}{2}\ \ & = \√6{2 \√{2}}。\ \ _ \广场结束{对齐}$

合理化的分母 $\ frac {\ sqrt {y-1}} {\ sqrt {y + 1}}$．

---

我们有

$\ begin {对齐} \ dfrac {\ sqrt {y-1}} {\ sqrt {y + 1}}＆= \ dfrac {\ big（\ sqrt {y-1} \ big）\ big（\ sqrt {y1} \大）} {\大（\ SQRT {Y + 1} \大）\大（\ SQRT {Y + 1} \大）} \\＆= \ dfrac {\ SQRT {Y ^ 2-1}} {Y + 1}。\ _ \ square \ END {对齐}$

合理化的表达式 $\ frac {y ^ 2 \ sqrt {x ^ 2 - 1}} {\ sqrt {x + 1}}。$

---

对分母进行理性化，我们得到

$\压裂{y ^ 2 \√{x ^ 2 - 1}} {\ sqrt {x + 1}} \ * \压裂{\ sqrt {x + 1}} {\ sqrt {x + 1}} = \压裂{y ^ 2 \√x ^ 2 {1} \ * \ sqrt {x + 1}} {x + 1}。$

进一步简化，我们有

${对齐}\ \开始压裂{y ^ 2 \√{(x - 1) (x + 1)} \ * \ sqrt {x + 1}} {x + 1} & = \压裂{y ^ 2 \√{(x - 1) (x + 1) ^ 2}} {x + 1} \ \ & = \压裂{y ^ 2 x + 1 \√{x}} {x + 1}。结束\{对齐}$

取消 $x + 1，$我们的最终结果是

$y ^ 2 \ sqrt {x-1}。\ _ \ square$

合理化的分母 $\ frac {3} {\ sqrt [3] {3}}$

---

我们有

$\ begin {对齐} \ dfrac {3} {\ sqrt [3] {3}}＆= \ dfrac {3×\ sqrt [3] {9}} {\ sqrt [3] {3}×\ sqrt [3]] {9}} \\＆= \ dfrac {3×\ SQRT [3] {9}} {\ SQRT [3] {27}} \\＆= \ dfrac {3×\ SQRT [3] {9}} {3} \\＆= \ sqrt [3] {9}。\ _ \ square \ END {对齐}$

现在考虑形式的分数 $\压裂{一}{b + \ sqrt {c}},$在哪里 $一种$那 $B.$和 $C$都是整数。如果我们乘以 $\ frac {b + \ sqrt {c}} {b + \ sqrt {c}}，$我们将获得 $\ frac {ab + a \ sqrt {c}} {b ^ 2 + 2b \ sqrt {c} + c}，$你可以看到，这是一个更复杂的分数形式。因此，当我们得到这类分数时，我们利用平方差公式 $(x - y) (x + y = x ^ 2 - y ^ 2。$这特别有用，因为如果 $x = \ sqrt {m}$和 $y = \ sqrt {n},$然后 $大\ \√大概{n} {m} - \ \大)大\ \√{m} + \ sqrt {n} \大)= m - n。$我们现在准备通过乘法来合理地利用原始表达的分母 $\压裂{b - \ sqrt {c}} {b - \ sqrt {c}}:$ $\压裂{一}{b + \ sqrt {c}} \ * \压裂{b - \ sqrt {c}} {b - \ sqrt {c}} = \压裂{ab - \ sqrt {c}} {b ^ 2 - c}。$

简化以下表达式:

$\ frac {3} {\ sqrt {3}} + \ frac {2} {\ sqrt {2} + \ sqrt {5}}。$

---

让我们分别处理这两个部分。 $\压裂{3}{\ sqrt {3}}$可以通过合理化来简化 $\压裂{\ sqrt {3}} {\ sqrt {3}}$和正确的部分 ${√{2}+√{5}}$可以通过乘以 $\ frac {\ sqrt {2} - \ sqrt {5}} {\ sqrt {2} - \ sqrt {5}}：$ $\压裂{3}{\ sqrt{3}} \ * \压裂{\ sqrt {3}} {\ sqrt{3}} + \压裂{2}{\ sqrt大概{5}}{2}+ \ \ * \压裂大概{2}- {\ \ sqrt {5}} {\ sqrt {2} - \ sqrt{5}} = \压裂{3 \ sqrt{3}}{3} + \压裂{2 \√{2}- \ sqrt{5})}{3}。$从这一点我们获得了答案 $\ frac {3 \ sqrt {3} -2 \ sqrt {2} +2 \ sqrt {5}} {3}。\ _ \ square$

简化以下表达式: $\ frac {7 \ sqrt {6} + \ sqrt {303}} {\ sqrt {3}} \ times（\ sqrt {101} - \ sqrt {98}）。$

---

我们从写作开始 $\ sqrt {98}$作为 $7 \ sqrt {2}:$ $\压裂{7 \ SQRT {6} + \ SQRT {303}} {\ SQRT {3}} \时间（\ SQRT {101} - 7 \ SQRT {2}）。$合理化分母 $\ sqrt {3}$给 $\开始{对齐}\压裂大概{6}{7 \ \ * \ sqrt {3} + \ sqrt {303} \ sqrt {3}} {3} \ * \ sqrt{101} - 7 \√{2})= & \压裂{7 \ sqrt{3 \ * 2 \乘以3}+ \ sqrt {3 \ times101 \乘以3}}{3}\ * \ sqrt{101} - 7 \√{2 } )\\ =& \ 压裂{7 \ sqrt {3 ^ 2 \ * 2} + \ sqrt {3 ^ 2 \ times101}} {3} \ * \ sqrt{101} - 7 \√{2 } )\\ =& \ 压裂大概{2}{3 \ * 7 \ +3大概{101}}{3}\ \ * \ sqrt{101} - 7 \√{2})。结束\{对齐}$取消了 $3.$它在左边 $\ begin {对齐}（7 \ sqrt {2} + \ sqrt {101}）\ times（\ sqrt {101} - 7 \ sqrt {2}）＆= 7 \ sqrt {202} - 98 + 101 - 7sqrt {202} \\＆= 3。\ _ \ square \ END {对齐}$

简化

$大\ dfrac{\{\√{5}\大)}^ 3 -{\大(大概{4}\ \大)}^ 3}{{\大(大概{5}\ \大)}^ 3 +{\大(大概{4}\ \大)}^ 3}。$

---

我们有

${对齐}\ \开始dfrac{{\√{5})}^ 3 - {(\ sqrt{4})} ^ 3}{{\√{5})}^ 3 +{\√{4})}^ 3}& = \ dfrac{5 - \ \√√6 4)5 + \√{20}+ 4)}{\√5 + \√4)5 - \√{20}+ 4 )}\\ & = \ dfrac{\ \√大大概{4}{5}- \ \大)、大(9 + 2 \√{5}\大)}{\大\√大概{4}{5}+ \ \大)大(9 - 2 \ \ sqrt{5} \大 )}\\ & = \ dfrac{\ \√大大概{4}{5}- \ \大)\大(9+ 2 \√{5}\大)\大(9 + 2 \√{5}\大)}{\大\√大概{4}{5}+ \ \大)\大9 - 2 \√{5}\大)大(9 + 2 \ \ sqrt{5} \大 )}\\ & = \ dfrac{\ \√大{5}- 2 \大)\大(81 + 20 + 36 \ sqrt{5} \大)}{\大\√{5}+ 2 \大)(81 - 20 )}\\ & = \ dfrac{\大(大概5 - 2 \ \大)\大(101 + 36 \ sqrt{5} \大)}{\大(大概5 + 2 \ \大)(61 )}\\ & = \ dfrac{\大(101 + 36 \ sqrt{5} \大)\大(大概5 - 2 \ \大)}{(61)\ \√5 + 2大\大 )}\\ & = \ dfrac{\大(101 + 36 \ sqrt{5} \大)、大(大概5 - 2 \ \大)\大(大概5 - 2 \ \大)}{(61)大\ \√5 + 2 \大)\ \√5 - 2大\大 )}\\ & = \ 大(101 + 36 dfrac {\ \ sqrt{5} \大)、大(5 + 4 - 4 \√5 \大)}{(61)(5 - 4 )}\\ & = \ 大(101 + 36 dfrac {\ \ sqrt{5} \大)\大大概5(9 - 4 \ \大)}{61}\ \ & = \ dfrac{189 - 80\根号5}{61}。\ _ \广场结束\{对齐}$

合理化的分母 $大概{\ dfrac {{\ \ sqrt [3] {2}}}} {{\ sqrt [4] {\ sqrt {5}}} - {\ sqrt [4] {\ sqrt{3}}}}。$

---

我们有

$\ begin {对齐} \ dfrac {{\ sqrt {\ sqrt [3] {\ sqrt [4] {\ sqrt {5}}} - {\ sqrt [4] {\ sqrt [4] {\ sqrt {3}}}＆= \ dfrac {\ left（2 ^ {\ frac {1} {3} \右）^ {\ frac {1} {2}} {{\ left（5 ^ {\ frac {右图）^ {\ frac {1} {2}} - \ left（3 ^ {\ frac {1} {4}}右）^ {\ frac {1} {2}}}} \\＆= \ dfrac {2 ^ {\ frac {1} {6}} {5 ^ {\ frac {1} {8}} - 3 ^ {\ frac {1} {8}}}\＆r = dfrac {2 ^ {\ frac {1} {6}}左（5 ^ {\ frac {1} {8}} + 3 ^ {\ frac {1} {8}}}{\ left（5 ^ {\ frac {1} {8}} - 3 ^ {\ frac {1} {8}} \ left）\ left（5 ^ {\ frac {1} {8}} + 3 ^{\ frac {1} {8} \右）} \\＆= \ dfrac {2 ^ {\ frac {1} {6}} \ left（5 ^ {\ frac {1} {8}} + 3^ {\ frac {1} {8}} \右）\ left（5 ^ {\ frac {1} {4}} + 3 ^ {\ frac {1} {4}}} {\ left（5 ^ {\ frac {1} {4}} - 3 ^ {\ frac {1} {4} \右）\ left（5 ^ {\ frac {1} {4}} + 3 ^ {\ frac {1}{4}}\right)}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right)}{\left(5^{\frac{1}{2}} - 3^{\frac{1}{2}}\right)}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{\left(5^{\frac{1}{2}} - 3^{\frac{1}{2}}\right)\left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{5-3}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{2}.\ _\square \end{aligned}$

合理化的分母 $\ dfrac {1} {\ SQRT [5] {\ SQRT [4] {\ SQRT [3] {\ SQRT {2}}}}}。$

---

我们有

$\ begin {对齐} \ dfrac {1} {\ sqrt [5] {\ sqrt [4] {\ sqrt [3] {\ sqrt {2}}}}＆= dfrac {1×{\ left（\sqrt [5] {\ sqrt [4] {\ sqrt [3] {\ sqrt {2}}}} \ otive）} \ excly）} ^ {4}} {\ sqrt [5] {\ sqrt [4] {\ sqrt [4] {\ sqrt [3] {\ sqrt {2}}}}×{\ left（\ sqrt [5] {\ sqrt [4] {\ sqrt [3] {\ sqrt {2}}}} \ rectle）} ^ {4}} \\＆= \ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt {2}}}} {\ sqrt [4] {\ sqrt [3] {\ sqrt {2}}}}}}\＆= dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt {2}}××{\ left（\ sqrt [4] {\ sqrt [3] {\ sqrt {2}}}右）} ^ {3}} {\ sqrt [4] {\ sqrt [3] {\ sqrt {2}}}×{\ left（\ sqrt [4] {\ sqrt [3] {\ sqrt {2}}} \右）} ^ {3}} \\＆= \ dfrac {\ SQRT [5] {\ SQRT [3] {\ SQRT {2}}}×\ SQRT [4] {\ SQRT {2}}{\ sqrt [3] {\ sqrt {2}}} \\＆= \ dfrac {\ sqrt [3] {\ sqrt [3] {\ sqrt {2}}}×\ sqrt [4] {\ sqrt [4] {\ sqrt{2}}×{\左（\ SQRT [3] {\ SQRT {2}} \右）} ^ {2}} {\ SQRT [3] {\ SQRT {2}}×{\左（\ SQRT[3] {\ sqrt {2}} \ =}} ^ {2}} \\＆= dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt {2}}××\ sqrt [4] {\ sqrt {2}}×\ sqrt [3] {2}} {\ sqrt {2}} \\＆= \ dfrac {\ sqrt [3] {\ sqrt [3] {\ sqrt {2}}}×\ sqrt [4] {\ sqrt {2}}×\ sqrt [3] {2}×\ sqrt {2}}} {\ sqrt {2}×\ sqrt {2}×\\＆= \ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt {2}}×\ sqrt [4] {\ sqrt {2}}×\ sqrt [3] {2}×\ sqrt {2}} {2}。\ _ \ square \ END {对齐}$

• 自由基方程
• 简化表达式