Biến đổi đơn giản biểu thức chứa căn

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: $\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$

$M = {\left( {\sqrt x  + \sqrt y } \right)^2} = {\left( {\sqrt x } \right)^2} + 2\sqrt x .\sqrt y  + {\left( {\sqrt y } \right)^2} = x + 2\sqrt {xy}  + y$

$N = \dfrac{{x\sqrt x  - y\sqrt y }}{{\sqrt x  - \sqrt y }} = \dfrac{{{{\left( {\sqrt x } \right)}^3} - {{\left( {\sqrt y } \right)}^3}}}{{\sqrt x  - \sqrt y }} = \dfrac{{\left( {\sqrt x  - \sqrt y } \right)\left( {x + \sqrt {xy}  + y} \right)}}{{\sqrt x  - \sqrt y }} = x + \sqrt {xy}  + y$

$P = \left( {\sqrt x  - \sqrt y } \right)\left( {\sqrt x  + \sqrt y } \right) = {\left( {\sqrt x } \right)^2} - {\left( {\sqrt y } \right)^2} = x - y$

Vậy $N = x + \sqrt {xy}  + y$.

Ta có: $\sqrt {9{x^2} - 16}  = 3\sqrt {3x - 4}$$\Leftrightarrow \sqrt {9{x^2} - 16}  = \sqrt {9\left( {3x - 4} \right)}  \Leftrightarrow \sqrt {9{x^2} - 16}  = \sqrt {27x - 36}$

Điều kiện: $27x - 36 \ge 0 \Leftrightarrow x \ge \dfrac{4}{3}$.

Với điều kiện trên ta có:

$\sqrt {9{x^2} - 16}  = \sqrt {27x - 36}  \Leftrightarrow 9{x^2} - 16 = 27x - 36 \Leftrightarrow 9{x^2} - 27x + 20 = 0 \Leftrightarrow 9{x^2} - 15x - 12x + 20 = 0$

$\Leftrightarrow 3x\left( {3x - 5} \right) - 4\left( {3x - 5} \right) = 0 \Leftrightarrow \left( {3x - 4} \right)\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}3x - 4 = 0\\3x - 5 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{4}{3}\\x = \dfrac{5}{3}\end{array} \right.\left( {TM} \right)$

Vậy phương trình đã cho có hai nghiệm phân biệt: $x = \dfrac{4}{3};x =  \dfrac{5}{3}$.

Điều kiện: $\left\{ \begin{array}{l}4x - 8 \ge 0\\9x - 18 \ge 0\\\dfrac{{x - 2}}{4} \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}4\left( {x - 2} \right) \ge 0\\9\left( {x - 2} \right) \ge 0\\x - 2 \ge 0\end{array} \right. \Leftrightarrow x - 2 \ge 0 \Leftrightarrow x \ge 2$

Ta có: $\sqrt {4x - 8}  - 2\sqrt {\dfrac{{x - 2}}{4}}  + \sqrt {9x - 18}  = 8$$\Leftrightarrow \sqrt {4\left( {x - 2} \right)}  - 2\sqrt {\dfrac{1}{4}.\left( {x - 2} \right)}  + \sqrt {9.\left( {x - 2} \right)}  = 8$

y$\Leftrightarrow 2\sqrt {x - 2}  - 2.\dfrac{1}{2}\sqrt {x - 2}  + 3\sqrt {x - 2}  = 8$

$\Leftrightarrow 2\sqrt {x - 2}  - \sqrt {x - 2}  + 3\sqrt {x - 2}  = 8$

$\Leftrightarrow 4\sqrt {x - 2}  = 8 \Leftrightarrow \sqrt {x - 2}  = 2 \Leftrightarrow x - 2 = 4 \Leftrightarrow x = 6$ (TM)

Vậy phương trình có một nghiệm $x = 6$.

Ta có: $\dfrac{{4a}}{{\sqrt 7  - \sqrt 3 }} - \dfrac{{2a}}{{2 - \sqrt 2 }} - \dfrac{a}{{\sqrt 3  + \sqrt 2 }}$$= \dfrac{{4a\left( {\sqrt 7  + \sqrt 3 } \right)}}{{\left( {\sqrt 7  - \sqrt 3 } \right)\left( {\sqrt 7  + \sqrt 3 } \right)}} - \dfrac{{2a\left( {2 + \sqrt 2 } \right)}}{{\left( {2 - \sqrt 2 } \right)\left( {2 + \sqrt 2 } \right)}} - \dfrac{{a\left( {\sqrt 3  - \sqrt 2 } \right)}}{{\left( {\sqrt 3  + \sqrt 2 } \right)\left( {\sqrt 3  - \sqrt 2 } \right)}}$

$= a\left( {\sqrt 7  + \sqrt 3 } \right) - a\left( {2 + \sqrt 2 } \right) - a\left( {\sqrt 3  - \sqrt 2 } \right)$$= a\left( {\sqrt 7  + \sqrt 3  - 2 - \sqrt 2  - \sqrt 3  + \sqrt 2 } \right)$

$= a\left( {\sqrt 7  - 2} \right)$

$\begin{array}{l}P = 3\sqrt {8x}  - 5\sqrt {48x}  + 9\sqrt {18x}  + 5\sqrt {12x} \\ = 3\sqrt {4.2.x}  - 5\sqrt {16.3.x}  + 9\sqrt {9.2.x}  + 5\sqrt {4.3.x} \\ = 3.2\sqrt {2x}  - 5.4\sqrt {3x}  + 9.3\sqrt {2x}  + 5.2\sqrt {3x} \\ = 6\sqrt {2x}  - 20\sqrt {3x}  + 27\sqrt {2x}  + 10\sqrt {3x} \\ = 33\sqrt {2x}  - 10\sqrt {3x} .\end{array}$

Điều kiện : $x \ge 3.$

$\begin{array}{l}\sqrt {{x^2} - 9}  - 3\sqrt {x - 3}  = 0\\ \Leftrightarrow \sqrt {\left( {x - 3} \right)\left( {x + 3} \right)}  - 3\sqrt {x - 3}  = 0\\ \Leftrightarrow \sqrt {x - 3} \left( {\sqrt {x + 3}  - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt {x - 3}  = 0\\\sqrt {x + 3}  - 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x - 3 = 0\\\sqrt {x + 3}  = 3\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 3\\x + 3 = 9\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\,\,\,\,\,\left( {tm} \right)\\x = 6\,\,\,\,\,\left( {tm} \right)\end{array} \right.\end{array}$

Điều kiện: $\left\{ \begin{array}{l}{x^2} - 4 \ge 0\\x + 2 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge 2\\x \le  - 2\end{array} \right.\\x \ge  - 2\end{array} \right.$$\Leftrightarrow x \ge 2$ hoặc $x=-2$

$\begin{array}{l}\sqrt {{x^2} - 4}  - 2\sqrt {x + 2}  = 0\\ \Leftrightarrow \sqrt {\left( {x - 2} \right)\left( {x + 2} \right)}  - 2\sqrt {x + 2}  = 0\\ \Leftrightarrow \sqrt {x + 2} \left( {\sqrt {x - 2}  - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt {x + 2}  = 0\\\sqrt {x - 2}  - 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x + 2 = 0\\\sqrt {x - 2}  = 2\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  - 2\\x - 2 = 4\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - 2\,\,\,\left( {tm} \right)\\x = 6\,\,\,\,\left( {tm} \right)\end{array} \right.\end{array}$

$\begin{array}{l}\,\,\,\,\,2\sqrt {8\sqrt 3 }  - 2\sqrt {5\sqrt 3 }  - 3\sqrt {20\sqrt 3 } \\ = 2\sqrt {4.2} .\sqrt {\sqrt 3 }  - 2\sqrt 5 .\sqrt {\sqrt 3 }  - 3\sqrt {4.5} .\sqrt {\sqrt 3 } \\ = 2.2\sqrt 2 \sqrt {\sqrt 3 }  - 2\sqrt 5 \sqrt {\sqrt 3 }  - 3.2.\sqrt 5 .\sqrt {\sqrt 3 } \\ = 4\sqrt {2\sqrt 3 }  - \left( {2 + 3.2} \right)\sqrt 5 \sqrt {\sqrt 3 } \\ = 4\sqrt {2\sqrt 3 }  - 8\sqrt {5\sqrt 3 } \end{array}$

Với các biểu thức $A,B,C$ mà $A,B,C > 0$, ta có: $\sqrt {\dfrac{A}{{BC}}}  = \dfrac{{\sqrt {ABC} }}{{\left| {BC} \right|}} = \dfrac{{\sqrt {ABC} }}{{BC}}$ (vì $B,C > 0$)

Với hai biểu thức $A,B$ mà $A,\,B \ge 0$, ta có: $\sqrt {{A^2}B}  = \left| A \right|\sqrt B  = A\sqrt B$

Ta có: $\sqrt {144{{\left( {3 + 2a} \right)}^4}}  = \sqrt {{{12}^2}.{{\left[ {{{\left( {3 + 2a} \right)}^2}} \right]}^2}}  = 12.\left| {{{\left( {3 + 2a} \right)}^2}} \right| = 12{\left( {3 + 2a} \right)^2}$

Ta có: $- 7x\sqrt {2xy}$ $=  - \sqrt {{{\left( {7x} \right)}^2}2xy}$$= -\sqrt {49{x^2}.2xy}  = -\sqrt {98{x^3}y}$.

Ta có: $5x\sqrt {\dfrac{{ - 12}}{{{x^3}}}}$$=  - \sqrt {{{\left( {5x} \right)}^2}.\dfrac{{ - 12}}{{{x^3}}}}  = \sqrt {25{x^2}\left( {\dfrac{{ - 12}}{x^3}} \right)}  =  - \sqrt {\dfrac{{ - 300}}{x}}$.

Ta có: $9\sqrt 7  = \sqrt {{9^2}.7}  = \sqrt {81.7}  = \sqrt {567}$; $8\sqrt 8  = \sqrt {{8^2}.8}  = \sqrt {64.8}  = \sqrt {512}$

Vì $512 < 567 \Leftrightarrow \sqrt {512}  < \sqrt {567}  \Leftrightarrow 8\sqrt 8  < 9\sqrt 7$

Vì $x < 0;y > 0$ nên ta có: $- 2{x^2}y\sqrt {\dfrac{{ - 9}}{{{x^3}{y^2}}}}$$=  - 2{x^2}y\dfrac{{\sqrt { - 9{x^3}{y^2}} }}{{\left| {{x^3}{y^2}} \right|}} =  - 2{x^2}y\dfrac{{\sqrt { - 9x.{x^2}} .\sqrt {{y^2}} }}{{\left( { - {x^3}{y^2}} \right)}} = 2.\dfrac{{\sqrt { - {3^2}x} .\left| x \right|.\left| y \right|}}{{xy}} = \dfrac{{2.3\sqrt { - x} \left( { - x} \right).y}}{{xy}} =  - 6\sqrt {-x}$.

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 $a = 7;b = 1 \Rightarrow a + b = 7 + 1 = 8$.

Ta có: $\sqrt {27x}  - \sqrt {48x}  + 4\sqrt {75x}  + \sqrt {243x}$$= \sqrt {9.3x}  - \sqrt {16.3x}  + 4\sqrt {25.3x}  + \sqrt {81.3x}$

$= \sqrt {{3^2}.3x}  - \sqrt {{4^2}.3x}  + 4\sqrt {{5^2}.3x}  + \sqrt {{{9}^2}.3x}$

$= 3\sqrt {3x}  - 4\sqrt {3x}  + 4.5\sqrt {3x}  + 9\sqrt {3x}  = \sqrt {3x} \left( {3 - 4 + 20+ 9} \right) = 28\sqrt {3x}$

Ta có: $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$