Liên hệ giữa phép chia và phép khai phương

Ta có $ \sqrt{\dfrac{{{a}^{4}}}{{{b}^{2}}}}=\dfrac{\sqrt{{{a}^{4}}}}{\sqrt{{{b}^{2}}}}=\dfrac{\sqrt{{{\left( {{a}^{2}} \right)}^{2}}}}{\sqrt{{{b}^{2}}}}=\dfrac{\left| {{a}^{2}} \right|}{\left| b \right|}=\dfrac{{{a}^{2}}}{\left| b \right|} $ .

$ D=\dfrac{2(a+b)}{\sqrt{b}}\sqrt{\dfrac{b}{{{a}^{2}}+2ab+{{b}^{2}}}}=\dfrac{2(a+b)}{\sqrt{b}}.\dfrac{\sqrt{b}}{\sqrt{{{a}^{2}}+2ab+{{b}^{2}}}}=\dfrac{2(a+b)}{\sqrt{b}}.\dfrac{\sqrt{b}}{\sqrt{{{(a+b)}^{2}}}} $

$ =\dfrac{2(a+b)}{\sqrt{b}}.\dfrac{\sqrt{b}}{|a+b|}=\dfrac{2(a+b)}{\sqrt{b}}.\dfrac{\sqrt{b}}{a+b}=2 $ (vì $ a,b > 0\Rightarrow a+b > 0\Rightarrow \left| a+b \right|=a+b $ ).

Ta có: $ \dfrac{3m}{8n}\sqrt{\dfrac{64{{n}^{2}}}{9{{m}^{2}}}}=\dfrac{3m}{8n}\sqrt{\dfrac{{{(8n)}^{2}}}{{{(3m)}^{2}}}}=\dfrac{3m}{8n}.\dfrac{|8n|}{|3m|}=\dfrac{3m.(-8n)}{8n.3m}=-1 $ (vì $ m > 0;n < 0) $ .

$ E=\dfrac{a-b}{2\sqrt{a}}\sqrt{\dfrac{ab}{{{(a-b)}^{2}}}}=\dfrac{a-b}{2\sqrt{a}}.\dfrac{\sqrt{ab}}{\sqrt{{{(a-b)}^{2}}}}=\dfrac{a-b}{2\sqrt{a}}.\dfrac{\sqrt{a}.\sqrt{b}}{|a-b|}=\dfrac{(a-b)\sqrt{b}}{2|a-b|} $

Mà $ 0 < a < b $ nên $ a-b < 0\Rightarrow |a-b|=-(a-b) $ . Khi đó $ E=\dfrac{(a-b)\sqrt{b}}{-2(a-b)}=\dfrac{-\sqrt{b}}{2} $

Ta có $ A=\sqrt{\dfrac{64{{x}^{3}}}{4x}}=\sqrt{16{{x}^{2}}}=\left| 4x \right|=4x $ do $ \left( x > 0 \right) $

$\begin{array}{*{20}{l}}{A = \left( {2\sqrt {18} - 3\sqrt {32} + 6\sqrt 2 } \right):\sqrt 2 }\\{ = \dfrac{{2\sqrt {18} }}{{\sqrt 2 }} - \dfrac{{3\sqrt {32} }}{{\sqrt 2 }} + \dfrac{{6\sqrt 2 }}{{\sqrt 2 }}}\\{ = 2.\sqrt {\dfrac{{18}}{2}} - 3.\sqrt {\dfrac{{32}}{2}} + 6}\\{ = 2.\sqrt 9 - 3.\sqrt {16} + 6}\\{ = 6 - 12 + 6 = 0}\end{array}$

$ \sqrt{\dfrac{1,21}{576}}=\dfrac{\sqrt{1,21}}{\sqrt{576}}=\dfrac{\sqrt{1,{{1}^{2}}}}{\sqrt{{{24}^{2}}}}=\dfrac{1,1}{24}=\dfrac{11}{240} $ .

Ta có $ \dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{{{a}^{3}}}+\sqrt{{{b}^{3}}}}{a-b}=\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\left( \sqrt{a}+\sqrt{b} \right)\left[ {{\left( \sqrt{a} \right)}^{2}}-\sqrt{a}.\sqrt{b}+{{\left( \sqrt{b} \right)}^{2}} \right]}{{{\left( \sqrt{a} \right)}^{2}}-{{\left( \sqrt{b} \right)}^{2}}} $

$ =\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\left( \sqrt{a}+\sqrt{b} \right)\left( a-\sqrt{ab}+b \right)}{\left( \sqrt{a}-\sqrt{b} \right)\left( \sqrt{a}+\sqrt{b} \right)}=\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}=\dfrac{a-b-a+\sqrt{ab}-b}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}-2b}{\sqrt{a}-\sqrt{b}} $

$ \sqrt{\dfrac{81}{169}}=\dfrac{\sqrt{81}}{\sqrt{169}}=\dfrac{\sqrt{{{9}^{2}}}}{\sqrt{{{13}^{2}}}}=\dfrac{9}{13} $ .

Ta có $ A=\dfrac{\sqrt{{{x}^{2}}-5x}}{\sqrt{x-5}} $ $ =\dfrac{\sqrt{x(x-5)}}{\sqrt{x-5}}=\dfrac{\sqrt{x}\sqrt{x-5}}{\sqrt{x-5}}=\sqrt{x} $

Để $ A=B $

$ \Leftrightarrow \sqrt{x}=x\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow \sqrt{x}\left( \sqrt{x}-1 \right)=0\Leftrightarrow \left[ \begin{array}{l} \sqrt{x}=0 \\ \sqrt{x}-1=0 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=0 \\ \sqrt{x}=1 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=0 \\ x=1 \end{array} \right. $ (loại vì $ x > 5 $ )

Vậy không có giá trị nào của $ x $ thỏa mãn điều kiện đề bài.

Điều kiện xác định của A là: $ \dfrac{x}{x-1}\ge 0\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x\ge 0 \\ x-1 > 0 \end{array} \right. \\ \left\{ \begin{array}{l} x\le 0 \\ x-1 < 0 \end{array} \right. \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x > 1 \\ x\le 0 \end{array} \right. $

Điều kiện xác định của B là: $ \left\{ \begin{array}{l} x\ge 0 \\ x-1 > 0 \end{array} \right.\Leftrightarrow x > 1 $

Khi đó với $ x > 1 $ thì cả 2 biểu thức đều xác định.

Đkxđ: $ \left\{ \begin{array}{l} x\ge \dfrac{1}{3} \\ x > 3 \end{array} \right.\Leftrightarrow x > 3 $

Ta có

$ \begin{array}{l} \dfrac{\sqrt{3x-1}}{\sqrt{x-3}}=2\Leftrightarrow \sqrt{\dfrac{3x-1}{x-3}}=2\Leftrightarrow \dfrac{3x-1}{x-3}=4 \\ \Leftrightarrow 3x-1=4x-12\Leftrightarrow x=11\Leftrightarrow x=11\left( t/m \right) \end{array} $

Vậy phương trình có nghiệm là $ x=11 $ .

Ta có

$ \begin{array}{l} A=\dfrac{a-\sqrt{b}}{\sqrt{b}}:\dfrac{\sqrt{b}}{a+\sqrt{b}} \\ =\dfrac{a-\sqrt{b}}{\sqrt{b}}.\dfrac{a+\sqrt{b}}{\sqrt{b}} \\ =\dfrac{\left( a-\sqrt{b} \right)\left( a+\sqrt{b} \right)}{\sqrt{b}.\sqrt{b}} \\ =\dfrac{{{a}^{2}}-b}{b} \end{array} $

$ 4{{a}^{4}}{{b}^{2}}.\sqrt{\dfrac{9}{{{a}^{8}}{{b}^{4}}}} $ $ =4{{a}^{4}}{{b}^{2}}.\dfrac{\sqrt{9}}{\sqrt{{{a}^{8}}{{b}^{4}}}}=4{{a}^{4}}{{b}^{2}}.\dfrac{3}{\sqrt{{{a}^{8}}}.\sqrt{{{b}^{4}}}} $ $ =\dfrac{12{{a}^{4}}{{b}^{2}}}{\sqrt{{{\left( {{a}^{4}} \right)}^{2}}}.\sqrt{{{\left( {{b}^{2}} \right)}^{2}}}}=\dfrac{12{{a}^{4}}{{b}^{2}}}{{{a}^{4}}.{{b}^{2}}}=12 $

Ta có $A = \frac{{\sqrt {27{x^4}{y^6}{z^2}} }}{{\sqrt {3{x^2}{y^2}} }} = \sqrt {\frac{{27{x^4}{y^6}{z^2}}}{{3{x^2}{y^2}}}} = \sqrt {9{x^2}{y^4}{z^2}} = 3\left| {xz} \right|{y^2} = - 3x{y^2}z$

$ \begin{array}{l} B=\dfrac{x-1}{\sqrt{y}-1}\dfrac{\sqrt{{{\left( \sqrt{y}-1 \right)}^{2}}}}{\sqrt{x-1}}=\dfrac{x-1}{\sqrt{y}-1}\dfrac{\left| \sqrt{y}-1 \right|}{\sqrt{x-1}} \\ =\dfrac{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+1 \right)}{\sqrt{x}-1}.\dfrac{1-\sqrt{y}}{\sqrt{y}-1}=-\left( \sqrt{x}+1 \right) \end{array} $

$\begin{array}{*{20}{l}}{A = \left( {x - 2} \right)\sqrt {\dfrac{{{x^2}}}{{4 - 4x + {x^2}}}} }\\{ = \left( {x - 2} \right)\sqrt {\dfrac{{{x^2}}}{{{{\left( {2 - x} \right)}^2}}}} }\\{ = \left( {x - 2} \right)\left| {\dfrac{x}{{2 - x}}} \right| = \left( {x - 2} \right)\left( {\dfrac{x}{{2 - x}}} \right) = - x}\end{array}$

Điều kiện: $ 2x-5 > 0\Leftrightarrow x > \dfrac{5}{2} $

Với điều kiện trên ta có: $ \dfrac{8+3x}{\sqrt{2x-5}}=\sqrt{2x-5} $ $ \Rightarrow 8+3x={{\left( \sqrt{2x-5} \right)}^{2}}\Leftrightarrow 8+3x=2x-5 $ $ \Leftrightarrow x=-13(KTM) $

Vậy phương trình vô nghiệm.

Ta có

$ \begin{array}{l} \sqrt{\dfrac{4}{9}}=\dfrac{\sqrt{4}}{\sqrt{9}}=\dfrac{2}{3} \\ \sqrt{\dfrac{9}{25}}=\dfrac{\sqrt{9}}{\sqrt{25}}=\dfrac{3}{5} \\ \sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{16+9}{16}}=\dfrac{\sqrt{25}}{\sqrt{16}}=\dfrac{5}{4} \\ \sqrt{\dfrac{25}{144}}=\dfrac{\sqrt{25}}{\sqrt{144}}=\dfrac{5}{12} \end{array} $

$ \sqrt{\dfrac{4}{9}}+\sqrt{\dfrac{9}{25}}+\sqrt{1\dfrac{9}{16}}+\sqrt{\dfrac{25}{144}}=\dfrac{2}{3}+\dfrac{3}{5}+\dfrac{5}{4}+\dfrac{5}{12}=\dfrac{44}{15} $

ĐKXĐ: $ x\ge 0;y\ge 0;x\ne y $ , khi đó ta có

$ \begin{array}{l} \dfrac{{{x}^{2}}-xy+y-x}{\sqrt{x}+\sqrt{y}}:\left( \sqrt{x}-\sqrt{y} \right)=0 \\ \Leftrightarrow \dfrac{{{x}^{2}}-xy+y-x}{\sqrt{x}+\sqrt{y}}.\dfrac{1}{\sqrt{x}-\sqrt{y}}=0 \\ \Leftrightarrow \dfrac{x\left( x-y \right)-\left( x-y \right)}{x-y}=0\Leftrightarrow \dfrac{\left( x-1 \right)\left( x-y \right)}{x-y}=0 \\ \Leftrightarrow x=1 \end{array} $

$ \dfrac{{{a}^{2}}}{11}.\dfrac{\sqrt{121}}{\sqrt{{{a}^{4}}}.\sqrt{{{b}^{10}}}}=\dfrac{{{a}^{2}}}{11}.\dfrac{\sqrt{{{11}^{2}}}}{\sqrt{{{\left( {{a}^{2}} \right)}^{2}}}.\sqrt{{{\left( {{b}^{5}} \right)}^{2}}}}=\dfrac{{{a}^{2}}}{11}.\dfrac{11}{{{a}^{2}}.\left| {{b}^{5}} \right|}=\dfrac{1}{\left| {{b}^{5}} \right|} $ .

Ta có $ A=\sqrt{\dfrac{x+3}{x-2}} $ xác định khi $ \dfrac{x+3}{x-2}\ge 0\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x+3\ge 0 \\ x-2 > 0 \end{array} \right. \\ \left\{ \begin{array}{l} x+3\le 0 \\ x-2 < 0 \end{array} \right. \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x > 2 \\ x\le -3 \end{array} \right. $

Ta có

$ \begin{array}{l} \dfrac{\sqrt{1100}}{\sqrt{11}}+\dfrac{\sqrt{5}}{\sqrt{20}}+\dfrac{\sqrt{112}}{\sqrt{7}} \\ =\sqrt{\dfrac{1100}{11}}+\sqrt{\dfrac{5}{20}}+\sqrt{\dfrac{112}{7}} \\ =\sqrt{100}+\sqrt{\dfrac{1}{4}}+\sqrt{16} \\ =10+\dfrac{1}{2}+4=\dfrac{29}{2} \end{array} $

Vì $ -729 < 0;625 > 0\Rightarrow \dfrac{625}{-729} < 0 $ nên không tồn tại căn bậc hai của số âm.

Ta có

$ \dfrac{\sqrt{9{{x}^{5}}+33{{x}^{4}}}}{\sqrt{3x+11}}=\dfrac{\sqrt{3{{x}^{4}}(3x+11)}}{\sqrt{3x+11}}=\dfrac{\sqrt{3}.\sqrt{{{x}^{4}}}.\sqrt{3x+11}}{\sqrt{3x+11}}=\sqrt{3}.\sqrt{{{\left( {{x}^{2}} \right)}^{2}}}=\sqrt{3}.\left| {{x}^{2}} \right|=\sqrt{3}{{x}^{2}} $