Với hai biểu thức $A, B$ mà $AB ≥ 0$ và $B ≠ 0$ ta có: \[\sqrt {\dfrac{A}{B}} = \dfrac{{\sqrt {A.B} }}{{\left| B \right|}}\]
Ví dụ:
\[\sqrt {\dfrac{{5a}}{{7b}}} = \sqrt {\dfrac{{5a.7b}}{{7b.7b}}} = \dfrac{{\sqrt {5a.7b} }}{{\sqrt {{{\left( {7b} \right)}^2}} }} = \dfrac{{\sqrt {35ab} }}{{7b}}\]
Với hai biểu thức $A, B$ mà $B > 0$ ta có: \[\dfrac{A}{{\sqrt B }} = \dfrac{{A\sqrt B }}{B}\]
Với các biểu thức $A, B, C$ mà $A ≥ 0$ và $A ≠ B^2$ ta có: \[\dfrac{C}{{\sqrt A \pm B}} = \dfrac{{C\left( {\sqrt A \pm B} \right)}}{{A - {B^2}}}\]
Với các biểu thức $A, B, C$ mà $A ≥ 0, B ≥ 0$ và $A ≠ B$ ta có: \[\dfrac{C}{{\sqrt A \pm \sqrt B }} = \dfrac{{C\left( {\sqrt A {\rm{ \;}} \pm B} \right)}}{{A - B}}\]
Ví dụ:
\[\dfrac{6}{{\sqrt 5 - \sqrt 3 }} = \dfrac{{6\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}} = \dfrac{{6\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} = 3\left( {\sqrt 5 + \sqrt 3 } \right)\]
Ta có
$ \begin{array}{l} \dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}=\dfrac{\sqrt{3}\left( \sqrt{\sqrt{3}+1}+1 \right)-\sqrt{3}\left( \sqrt{\sqrt{3}+1}-1 \right)}{\left( \sqrt{\sqrt{3}+1}+1 \right)\left( \sqrt{\sqrt{3}+1}-1 \right)} \\ =\dfrac{\sqrt{3\left( \sqrt{3}+1 \right)}+\sqrt{3}-\sqrt{3\left( \sqrt{3}+1 \right)}+\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{2\sqrt{3}}{\sqrt{3}}=2 \end{array} $
$ \begin{array}{l} \dfrac{1}{\sqrt{6}+\sqrt{5}+1}=\dfrac{\sqrt{6}-\left( \sqrt{5}+1 \right)}{\left( \sqrt{6}+\sqrt{5}+1 \right)\left( \sqrt{6}-\left( \sqrt{5}+1 \right) \right)} \\ =\dfrac{\sqrt{6}-\sqrt{5}-1}{6-{{\left( \sqrt{5}+1 \right)}^{2}}}=\dfrac{\sqrt{6}-\sqrt{5}-1}{-2\sqrt{5}}=\dfrac{\sqrt{6.5}-5-\sqrt{5}}{-2.5} \\ =\dfrac{\sqrt{5}+5-\sqrt{30}}{10} \end{array} $
Ta có: $ \dfrac{2}{3-\sqrt{5}}=\dfrac{2(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}=\dfrac{2(3+\sqrt{5})}{9-5}=\dfrac{3+\sqrt{5}}{2} $
$ \dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left( 2\sqrt{a}+\sqrt{b} \right)}{\left( 2\sqrt{a}-\sqrt{b} \right)\left( 2\sqrt{a}+\sqrt{b} \right)}=\dfrac{6a\left( 2\sqrt{a}+\sqrt{b} \right)}{4a-b} $
Ta có $ 5\sqrt{3}=\sqrt{{{5}^{2}}.3}=\sqrt{25.3}=\sqrt{75} $
$ 4\sqrt{5}=\sqrt{{{4}^{2}}.5}=\sqrt{16.5}=\sqrt{80} $
Vì $ 75 < 80\Leftrightarrow \sqrt{75} < \sqrt{80}\Leftrightarrow 5\sqrt{3} < 4\sqrt{5} $
Ta có $ 5\sqrt{a}+6\sqrt{\dfrac{a}{4}}-a\sqrt{\dfrac{4}{a}}+5\sqrt{\dfrac{4a}{25}} $
$ =5\sqrt{a}+6\sqrt{\dfrac{1}{4}.a}-a\sqrt{4.\dfrac{1}{a}}+5\sqrt{\dfrac{4}{25}.a} $
$ =5\sqrt{a}+6\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}.a}-a\sqrt{{{2}^{2}}.\dfrac{1}{a}}+5\sqrt{{{\left( \dfrac{2}{5} \right)}^{2}}.a} $
$ =5\sqrt{a}+6.\dfrac{1}{2}\sqrt{a}-2a\sqrt{\dfrac{1}{a}}+5.\dfrac{2}{5}\sqrt{a} $
$ =5\sqrt{a}+3\sqrt{a}-2a\dfrac{\sqrt{a}}{a}+2\sqrt{a}=5\sqrt{a}+3\sqrt{a}-2\sqrt{a}+2\sqrt{a}=8\sqrt{a} $ .
Ta có
$ \begin{array}{l} \dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\left( \sqrt{n+1}+\sqrt{n} \right)\left( \sqrt{n+1}-\sqrt{n} \right)} \\ =\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n} \end{array} $
Ta có
$ \begin{array}{l} \dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{{{x}^{3}}}-\sqrt{{{y}^{3}}}}{\sqrt{x}-\sqrt{y}} \\ =\dfrac{\left( \sqrt{x}-\sqrt{y} \right)\left( x+\sqrt{xy}+y \right)}{\sqrt{x}-\sqrt{y}} \\ =x+\sqrt{xy}+y \end{array} $
$ \begin{array}{l} \dfrac{2}{\sqrt{7}+\sqrt{5}}+\sqrt{5}=\dfrac{2\left( \sqrt{7}-\sqrt{5} \right)}{\left( \sqrt{7}+\sqrt{5} \right)\left( \sqrt{7}-\sqrt{5} \right)}+\sqrt{5} \\ =\dfrac{2\left( \sqrt{7}-\sqrt{5} \right)}{7-5}+\sqrt{5}=\sqrt{7}-\sqrt{5}+\sqrt{5}=\sqrt{7} \end{array} $
Vì $ x > 0;y > 0 $ nên $ xy > 0 $ . Từ đó ta có $ xy\sqrt{\dfrac{4}{{{x}^{2}}{{y}^{2}}}}=xy.\dfrac{\sqrt{4}}{\sqrt{{{x}^{2}}{{y}^{2}}}}=xy.\dfrac{2}{xy}=2 $
Ta có
$ \begin{array}{l} \dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{4}} \\ =\dfrac{\sqrt{2}-\sqrt{1}}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+\dfrac{\sqrt{4}-\sqrt{3}}{4-3}+\dfrac{\sqrt{5}-\sqrt{4}}{5-4} \\ =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4} \\ =\sqrt{5}-1 \end{array} $
Ta có $ \sqrt{\dfrac{4}{3}}+\sqrt{\dfrac{5}{27}}=\dfrac{2\sqrt{3}}{3}+\dfrac{\sqrt{15}}{9}=\dfrac{6\sqrt{3}+\sqrt{15}}{9} $
Ta có
$ \dfrac{1}{5+3\sqrt{2}}+\dfrac{1}{5-3\sqrt{2}}=\dfrac{5-3\sqrt{2}}{\left( 5+3\sqrt{2} \right)\left( 5-3\sqrt{2} \right)}+\dfrac{5+3\sqrt{2}}{\left( 5+3\sqrt{2} \right)\left( 5-3\sqrt{2} \right)}=\dfrac{10}{{{5}^{2}}-{{\left( 3\sqrt{2} \right)}^{2}}}=\dfrac{10}{25-18}=\dfrac{10}{7} $
Suy ra $ a=10;b=7\Rightarrow 2a=2.10=20 $ .
Ta có
$ \dfrac{4}{3\sqrt{x}+2\sqrt{y}}=\dfrac{4\left( 3\sqrt{x}-2\sqrt{y} \right)}{\left( 3\sqrt{x}+2\sqrt{y} \right)\left( 3\sqrt{x}-2\sqrt{y} \right)}=\dfrac{4\left( 3\sqrt{x}-2\sqrt{y} \right)}{{{\left( 3\sqrt{x} \right)}^{2}}-{{\left( 2\sqrt{y} \right)}^{2}}}=\dfrac{12\sqrt{x}-8\sqrt{y}}{9x-4y} $
Ta có:
$ \dfrac{2}{7+3\sqrt{5}}+\dfrac{2}{7-3\sqrt{5}}=\dfrac{2\left( 7-3\sqrt{5} \right)}{\left( 7+3\sqrt{5} \right)\left( 7-3\sqrt{5} \right)}+\dfrac{2\left( 7+3\sqrt{5} \right)}{\left( 7-3\sqrt{5} \right)\left( 7+3\sqrt{5} \right)} $
$ =\dfrac{14-6\sqrt{5}}{{{7}^{2}}-{{\left( 3\sqrt{5} \right)}^{2}}}+\dfrac{14+6\sqrt{5}}{{{7}^{2}}-{{\left( 3\sqrt{5} \right)}^{2}}}=\dfrac{14-6\sqrt{5}+14+6\sqrt{5}}{49-9.5}=\dfrac{28}{4}=\dfrac{7}{1} $
Suy ra $ a=7;b=1\Rightarrow a+b=7+1=8 $ .
$ 7\sqrt{x}+11y\sqrt{36{{x}^{5}}}-2{{x}^{2}}\sqrt{16x{{y}^{2}}}-\sqrt{25x} $ $ =7\sqrt{x}+11y\sqrt{{{6}^{2}}{{x}^{4}}.x}-2{{x}^{2}}\sqrt{{{4}^{2}}x{{y}^{2}}}-\sqrt{{{5}^{2}}x} $ $ =7\sqrt{x}+11y.6{{x}^{2}}\sqrt{x}-2{{x}^{2}}.4.y\sqrt{x}-5\sqrt{x} $ $ =7\sqrt{x}+66{{x}^{2}}y\sqrt{x}-8{{x}^{2}}y\sqrt{x}-5\sqrt{x} $
$ =\left( 7\sqrt{x}-5\sqrt{x} \right)+\left( 66{{x}^{2}}y\sqrt{x}-8{{x}^{2}}y\sqrt{x} \right)=2\sqrt{x}+58{{x}^{2}}y\sqrt{x} $ .
Ta có $ \dfrac{a}{\sqrt{5}+1}+\dfrac{a}{\sqrt{5}-2}-\dfrac{a}{3-\sqrt{5}}-\sqrt{5}a $
$ =\dfrac{a\left( \sqrt{5}-1 \right)}{\left( \sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)}+\dfrac{a\left( \sqrt{5}+2 \right)}{\left( \sqrt{5}-2 \right)\left( \sqrt{5}+2 \right)}-\dfrac{a\left( 3+\sqrt{5} \right)}{\left( 3+\sqrt{5} \right)\left( 3-\sqrt{5} \right)}-\sqrt{5}a $
$ =\dfrac{a\left( \sqrt{5}-1 \right)}{4}+\dfrac{a\left( \sqrt{5}+2 \right)}{1}-\dfrac{a\left( 3+\sqrt{5} \right)}{4}-\sqrt{5}a $
$ =\dfrac{a\left( \sqrt{5}-1 \right)+4a\left( 2+\sqrt{5} \right)-a\left( 3+\sqrt{5} \right)-4\sqrt{5}a}{4} $
$ =\dfrac{a\left( \sqrt{5}-1+8+4\sqrt{5}-3-\sqrt{5}-4\sqrt{5} \right)}{4}=\dfrac{4a}{4}=a $
Ta có
$ \begin{array}{l} \dfrac{a-\sqrt{a}}{1-\sqrt{a}}+\dfrac{1}{\sqrt{a}}=\dfrac{\sqrt{a}\left( \sqrt{a}-1 \right)}{1-\sqrt{a}}+\dfrac{1}{\sqrt{a}} \\ =-\sqrt{a}+\dfrac{1}{\sqrt{a}}=-\sqrt{a}+\dfrac{\sqrt{a}}{a}=\sqrt{a}\left( \dfrac{1}{a}-1 \right) \end{array} $
Ta có
$ \dfrac{1}{2+\sqrt{x}}-\dfrac{1}{2-\sqrt{x}}=\dfrac{2-\sqrt{x}-\left( 2+\sqrt{x} \right)}{\left( 2+\sqrt{x} \right)\left( 2-\sqrt{x} \right)}=\dfrac{-2\sqrt{x}}{4-x} $
Vì $ x < 0;y < 0 $ nên $ xy > 0 $ . Từ đó ta có
$ -xy\sqrt{\dfrac{3}{xy}}=-xy.\dfrac{\sqrt{3xy}}{xy}=-\sqrt{3xy} $ .
Ta có
$ \dfrac{3}{6+\sqrt{3a}}=\dfrac{3\left( 6-\sqrt{3a} \right)}{\left( 6+\sqrt{3a} \right)\left( 6-\sqrt{3a} \right)}=\dfrac{3\left( 6-\sqrt{3a} \right)}{{{6}^{2}}-{{\left( \sqrt{3a} \right)}^{2}}}=\dfrac{3\left( 6-\sqrt{3a} \right)}{36-3a}=\dfrac{6-\sqrt{3a}}{12-a} $ .
Ta có
$ \begin{array}{*{35}{l}} \begin{array}{l} \dfrac{5}{12\left( 2\sqrt{5}+3\sqrt{2} \right)}-\dfrac{5}{12\left( 2\sqrt{5}-3\sqrt{2} \right)} \\ =\dfrac{5\left( 2\sqrt{5}-3\sqrt{2} \right)-5\left( 2\sqrt{5}+3\sqrt{2} \right)}{12\left( 2\sqrt{5}+3\sqrt{2} \right)\left( 2\sqrt{5}-3\sqrt{2} \right)} \end{array} \\ =\dfrac{10\sqrt{5}-15\sqrt{2}-10\sqrt{5}-15\sqrt{2}}{12.2}=-\dfrac{5\sqrt{2}}{4} \end{array} $
Ta có $ \dfrac{2a}{2-\sqrt{a}}=\dfrac{2a\left( 2+\sqrt{a} \right)}{\left( 2-\sqrt{a} \right)\left( 2+\sqrt{a} \right)}=\dfrac{2a\sqrt{a}+4a}{4-a}. $
Ta có $ \left( \dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-1} \right):\dfrac{1}{2\sqrt{5}-\sqrt{6}} $
$ =\left( \dfrac{\sqrt{100}-\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{5}.\sqrt{6}-\sqrt{5}}{\sqrt{5}-1} \right):\dfrac{1}{2\sqrt{5}-\sqrt{6}} $ $ =\left( \dfrac{\sqrt{20}\left( \sqrt{5}+\sqrt{2} \right)}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{6}.\left( \sqrt{5}-1 \right)}{\sqrt{5}-1} \right):\dfrac{1}{2\sqrt{5}-\sqrt{6}} $ $ =\left( 2\sqrt{5}+\sqrt{6} \right)\left( 2\sqrt{5}-\sqrt{6} \right)={{\left( 2\sqrt{5} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}=20-6=14 $
Ta có
$ \begin{array}{l} A=\dfrac{\sqrt{1}+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5} \\ +\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9} \\ A=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5} \\ -\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}\\ A=-1-\sqrt{8}=-1-2\sqrt{2} \end{array} $
Ta có $ \dfrac{6}{\sqrt{x}+\sqrt{2y}} $ $ =\dfrac{6\left( \sqrt{x}-\sqrt{2y} \right)}{\left( \sqrt{x}+\sqrt{2y} \right)\left( \sqrt{x}-\sqrt{2y} \right)}=\dfrac{6\left( \sqrt{x}-\sqrt{2y} \right)}{x-2y} $ .
Ta có
$ \sqrt{\dfrac{3}{20}}+\sqrt{\dfrac{1}{60}}-2\sqrt{\dfrac{1}{15}}=\dfrac{\sqrt{3.20}}{20}+\dfrac{\sqrt{60}}{60}-\dfrac{2\sqrt{15}}{15}=\dfrac{3\sqrt{60}+\sqrt{60}-4.\sqrt{4.15}}{60}=\dfrac{4\sqrt{60}-4\sqrt{60}}{60}=0. $
Ta có $\frac{5}{{5 - \sqrt 3 }} = \frac{{5\left( {5 + \sqrt 3 } \right)}}{{{5^2} - {{\left( {\sqrt 3 } \right)}^2}}} = \frac{{5\left( {5 + \sqrt 3 } \right)}}{{25 - 3}} = \frac{{25 + 5\sqrt 3 }}{{22}}$
$ \Rightarrow \dfrac{5}{5-\sqrt{3}}-\dfrac{5\sqrt{3}}{22}=\dfrac{25}{22} $
Điều kiện: $ \left\{ \begin{array}{l} a\ge 0 \\ b\ge 0 \\ ab\ne 1 \end{array} \right. $
Ta có
$ \dfrac{\sqrt{a}+1}{\sqrt{ab}+1}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1=\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+a\sqrt{b}+\sqrt{ab}+\sqrt{a}-ab+1}{ab-1} $
$ =\dfrac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}=\dfrac{2\sqrt{ab}\left( \sqrt{a}+1 \right)}{ab-1}. $
Và
$ \dfrac{\sqrt{a}+1}{\sqrt{ab}+1}-\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1=\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-a\sqrt{b}-\sqrt{ab}-\sqrt{a}+ab-1}{ab-1} $
$ =\dfrac{-2\sqrt{a}-2}{ab-1}=\dfrac{-2\left( \sqrt{a}+1 \right)}{ab-1} $
Nên $ C=\dfrac{2\sqrt{ab}\left( \sqrt{a}+1 \right)}{ab-1}:\dfrac{-2\left( \sqrt{a}+1 \right)}{ab-1}=-\sqrt{ab}. $
Ta có: $ \dfrac{3}{2}\sqrt{6}+2.\sqrt{\dfrac{2}{3}}-4.\sqrt{\dfrac{3}{2}}=\dfrac{3}{2}\sqrt{6}+2.\dfrac{\sqrt{6}}{3}-4.\dfrac{\sqrt{6}}{2}=\sqrt{6}\left( \dfrac{3}{2}+\dfrac{2}{3}-2 \right)=\dfrac{\sqrt{6}}{6} $ .