Rút gọn biểu thức chứa căn

Điều kiện: $x \ge 0,\,\,x \ne 1.$ 

$\begin{array}{l}A = \dfrac{{x + \sqrt x  + 1}}{{x + \sqrt x  - 2}} + \dfrac{1}{{\sqrt x  - 1}} + \dfrac{1}{{\sqrt x  + 2}}\,\,\\\,\,\,\, = \dfrac{{x + \sqrt x  + 1 + \sqrt x  + 2 + \sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\, = \dfrac{{x + 3\sqrt x  + 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\, = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}} = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}.\end{array}$

$\left( {3\sqrt 2  + \sqrt 6 } \right)\sqrt {6 - 3\sqrt 3 }$$= \left( {3 + \sqrt 3 } \right).\sqrt 2 .\sqrt {6 - 3\sqrt 3 }  = \left( {3 + \sqrt 3 } \right).\sqrt {2\left( {6 - 3\sqrt 3 } \right)}$$= \left( {3 + \sqrt 3 } \right)\sqrt {12 - 2.3\sqrt 3 }  = \left( {3 + \sqrt 3 } \right)\sqrt {9 - 2.3.\sqrt 3  + 3}  = \left( {3 + \sqrt 3 } \right)\sqrt {{{\left( {3 - \sqrt 3 } \right)}^2}}$

$= \left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right) = 9 - 3 = 6.$

ĐKXĐ: $x \ge 0,\,\,x \ne 4.$

$Q = \dfrac{{2x - 3\sqrt x  - 2}}{{\sqrt x  - 2}}$$= \dfrac{{\left( {2\sqrt x  + 1} \right)\left( {\sqrt x  - 2} \right)}}{{\sqrt x  - 2}}$$= 2\sqrt x  + 1.$

Ta có: $x = 2020 - 2\sqrt {2019}$$= 2019 - 2\sqrt {2019}  + 1$$= {\left( {\sqrt {2019}  - 1} \right)^2}\,\,\,\left( {tm} \right)$

$\Rightarrow \sqrt x  = \sqrt {{{\left( {\sqrt {2019}  - 1} \right)}^2}}  = \left| {\sqrt {2019}  - 1} \right|$$= \sqrt {2019}  - 1\,.$

Thay $\sqrt x  = \sqrt {2019}  - 1$ vào biểu thức $Q$ ta được:

$Q = 2\left( {\sqrt {2019}  - 1} \right) + 1$$= 2\sqrt {2019}  - 2 + 1$$= 2\sqrt {2019}  - 1.$

Vậy $x = 2020 - 2\sqrt {2019}$ thì $Q = 2\sqrt {2019}  - 1.$

$\left( {\sqrt 5  - 1} \right)\sqrt {6 + 2\sqrt 5 }$$= \left( {\sqrt 5  - 1} \right)\sqrt {5 + 2\sqrt 5 .1 + 1}  = \left( {\sqrt 5  - 1} \right)\sqrt {{{\left( {\sqrt 5  + 1} \right)}^2}}  = \left( {\sqrt 5  - 1} \right)\left( {\sqrt 5  + 1} \right) = 5 - 1 = 4$

Điều kiện: $x \ge 0.$

$\begin{array}{l}P = \left( {\dfrac{{3\sqrt x }}{{x\sqrt x  + 1}} - \dfrac{{\sqrt x }}{{x - \sqrt x  + 1}} + \dfrac{1}{{\sqrt x  + 1}}} \right):\dfrac{{\sqrt x  + 3}}{{x - \sqrt x  + 1}}\\\,\,\,\, = \left[ {\dfrac{{3\sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}} - \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}} + \dfrac{{x - \sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}} \right]:\dfrac{{\sqrt x  + 3}}{{x - \sqrt x  + 1}}\\\,\,\,\, = \dfrac{{3\sqrt x  - x - \sqrt x  + x - \sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}.\dfrac{{x - \sqrt x  + 1}}{{\sqrt x  + 3}}\\\,\,\,\, = \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}.\dfrac{{x - \sqrt x  + 1}}{{\sqrt x  + 3}} = \dfrac{1}{{\sqrt x  + 3}}.\end{array}$

$\begin{array}{l} \Rightarrow P \ge \dfrac{1}{5} \Leftrightarrow \dfrac{1}{{\sqrt x  + 3}} \ge \dfrac{1}{5}\\ \Leftrightarrow \dfrac{1}{{\sqrt x  + 3}} - \dfrac{1}{5} \ge 0 \Leftrightarrow \dfrac{{5 - \sqrt x  - 3}}{{5\left( {\sqrt x  + 3} \right)}} \ge 0\\ \Leftrightarrow \dfrac{{2 - \sqrt x }}{{5\left( {\sqrt x  + 3} \right)}} \ge 0 \Leftrightarrow 2 - \sqrt x  \ge 0\\ \Leftrightarrow \sqrt x  \le 2 \Leftrightarrow x \le 4\end{array}$

Vậy $0 \le x \le 4$ thỏa mãn bài toán.

Điều kiện: $x \ge 0,\,\,x \ne 9.$

$\begin{array}{l}A + B = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\sqrt x \left( {\sqrt x  - 3} \right) + 2\sqrt x \left( {\sqrt x  + 3} \right) - 3x - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{x - 3\sqrt x  + 2x + 6\sqrt x  - 3x - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{3\sqrt x  - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{3\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{3}{{\sqrt x  + 3}}\end{array}$

Vậy $A + B = \dfrac{3}{{\sqrt x  + 3}}$ (với $x \ge 0,\,\,x \ne 9$).

Điều kiện: $x \ge 0,\,\,x \ne 9.$

Thay $x = 16$ (thỏa mãn điều kiện) vào biểu thức $A$ ta có: 

$A = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} = \dfrac{{\sqrt {16} }}{{\sqrt {16}  + 3}} = \dfrac{4}{{4 + 3}} = \dfrac{4}{7}$.

Vậy khi $x = 16$ thì $A = \dfrac{4}{7}$.

Ta có:

$\left( {\frac{1}{2}\sqrt {\frac{a}{2}} {\rm{\;}} - \frac{3}{2}\sqrt {2a} {\rm{\;}} + \frac{4}{5}\sqrt {200a} } \right):\frac{1}{8}$$= \left( {\frac{{\sqrt {2a} }}{4} - \frac{3}{2}\sqrt 2 a.{\rm{\;}} + \frac{4}{5}\sqrt {100} .\sqrt {2a} } \right).8$

$= 2\sqrt {2a} {\rm{\;}} - 12\sqrt {2a} {\rm{\;}} + 64\sqrt {2a} {\rm{\;}} = 54\sqrt {2a}$

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

Ta có $\left( {\dfrac{{\sqrt {14}  - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15}  - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{a\left( {\sqrt 7  - \sqrt 5 } \right)}}$$= \left( {\dfrac{{\sqrt 7 .\sqrt 2  - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 .\sqrt 3  - \sqrt 5 }}{{1 - \sqrt 3 }}} \right).a\left( {\sqrt 7  - \sqrt 5 } \right)$$= \left( {\dfrac{{ - \sqrt 7 \left( {1 - \sqrt 2 } \right)}}{{1 - \sqrt 2 }} - \dfrac{{\sqrt 5 \left( {\sqrt 3  - 1} \right)}}{{\sqrt 3  - 1}}} \right).a\left( {\sqrt 7  - \sqrt 5 } \right)$

$= \left[ {\dfrac{{\sqrt 6 \left( {\sqrt 2  - 1} \right)}}{{2\left( {\sqrt 2  - 1} \right)}} - 2\sqrt 6 } \right].\left( { - \dfrac{a}{{\sqrt 6 }}} \right)$$= \left( { - \sqrt 7  - \sqrt 5 } \right).a\left( {\sqrt 7  - \sqrt 5 } \right) =  - a.\left( {\sqrt 7  + \sqrt 5 } \right)\left( {\sqrt 7  - \sqrt 5 } \right) =  - 2a.$

Thay $x = 4$ (thỏa mãn điều kiện) vào $P$ ta được $P = \dfrac{{\sqrt 4 }}{{\sqrt 4  - 1}} = \dfrac{2}{{2 - 1}} = 2$.

Ta có: $x = \dfrac{8}{{3 - \sqrt 5 }} = \dfrac{{8\left( {3 + \sqrt 5 } \right)}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}} = \dfrac{{8\left( {3 + \sqrt 5 } \right)}}{{9 - 5}} = 6 + 2\sqrt 5  = {\left( {\sqrt 5  + 1} \right)^2}\left( {tm} \right)$$\Rightarrow \sqrt x  = \sqrt {{{\left( {\sqrt 5  + 1} \right)}^2}}  = \sqrt 5  + 1$

Khi đó ta có: $P = \dfrac{{\sqrt 5  + 1}}{{\sqrt 5  + 1 - 1}} = \dfrac{{\sqrt 5  + 1}}{{\sqrt 5 }} = \dfrac{{5 + \sqrt 5 }}{5}$.

Ta có: ${x^2} - 5x + 4 = 0 \Leftrightarrow {x^2} - 4x - x + 4 = 0 \Leftrightarrow x\left( {x - 4} \right) - \left( {x - 4} \right) = 0$

$\Leftrightarrow \left( {x - 1} \right)\left( {x - 4} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x - 1 = 0\\x - 4 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\left( {tm} \right)\\x = 4\left( {ktm} \right)\end{array} \right.$

Thay $x = 1$ vào biểu thức $P$ ta được $P = \dfrac{{\sqrt 1 }}{{\sqrt 1  - 2}} = \dfrac{1}{{ - 1}} =  - 1$.

Ta có $A = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x  + 2}} + \dfrac{{2 + 5\sqrt x }}{{4 - x}}$$= \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right) + 2\sqrt x \left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} - \dfrac{{2 + 5\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}$

$= \dfrac{{x + 3\sqrt x  + 2 + 2x - 4\sqrt x  - 2 - 5\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}$$= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}$$= \dfrac{{3\sqrt x \left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}} = \dfrac{{3\sqrt x }}{{\sqrt x  + 2}}$

Vậy $A = \dfrac{{3\sqrt x }}{{\sqrt x  + 2}}$ với $x \ge 0;x \ne 4$

Ta có $B = \left( {\dfrac{{\sqrt x  - 2}}{{x - 1}} - \dfrac{{\sqrt x  + 2}}{{x + 2\sqrt x  + 1}}} \right).\dfrac{{{{\left( {1 - x} \right)}^2}}}{2}$$= \left( {\dfrac{{\sqrt x  - 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}} - \dfrac{{\sqrt x  + 2}}{{{{\left( {\sqrt x  + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}$

$= \left( {\dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right){{\left( {\sqrt x  + 1} \right)}^2}}} - \dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right){{\left( {\sqrt x  + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}{{\left( {\sqrt x  + 1} \right)}^2}}}{2}$

$= \dfrac{{x - \sqrt x  - 2 - x - \sqrt x  + 2}}{{\left( {\sqrt x  - 1} \right){{\left( {\sqrt x  + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}.{{\left( {\sqrt x  + 1} \right)}^2}}}{2}$$= \dfrac{{ - 2\sqrt x \left( {\sqrt x  - 1} \right)}}{2} = \sqrt x  - x$

Vậy $B = \sqrt x  - x$.

Ta có $x - 5\sqrt x  + 6 = x - 2\sqrt x  - 3\sqrt x  + 6 = \sqrt x \left( {\sqrt x  - 2} \right) - 3\left( {\sqrt x  - 2} \right) = \left( {\sqrt x  - 3} \right)\left( {\sqrt x  - 2} \right)$ nên

$C = \dfrac{{2\sqrt x  - 9}}{{x - 5\sqrt x  + 6}} - \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 2}} - \dfrac{{2\sqrt x  + 1}}{{3 - \sqrt x }}$$= \dfrac{{2\sqrt x  - 9}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}} - \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 2}} + \dfrac{{2\sqrt x  + 1}}{{\sqrt x  - 3}}$

$= \dfrac{{2\sqrt x  - 9 - \left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right) + \left( {2\sqrt x  + 1} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}}$$= \dfrac{{2\sqrt x  - 9 - x + 9 + 2x - 3\sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}}$

$= \dfrac{{x - \sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}} = \dfrac{{x - 2\sqrt x  + \sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}} = \dfrac{{\sqrt x \left( {\sqrt x  - 2} \right) + \left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}} = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}} = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}$

Vậy $C = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}$với $x \ge 0;x \ne 4;x \ne 9$

$\forall x > 9:m\left( {\sqrt x  - 3} \right)P > x + 1 \Leftrightarrow m\left( {\sqrt x  - 3} \right).\dfrac{{4x}}{{\sqrt x  - 3}} > x + 1 \Leftrightarrow m.4x > x + 1 \Leftrightarrow m > \dfrac{{x + 1}}{{4x}}$
Ta có: với mọi giá trị x > 9 thì x + 1 > 9 + 1 = 10

4x > 4.9 = 36

Vậy $m > \dfrac{{10}}{{36}} = \dfrac{5}{{18}}$

Với điều kiện: $x > 0,x \ne 4,x \ne 9$ . Ta có: P = -1

$\begin{array}{l} \Leftrightarrow \dfrac{{4x}}{{\sqrt x  - 3}} =  - 1 \Leftrightarrow 4x + \sqrt x  - 3 = 0 \Leftrightarrow 4x + 4\sqrt x  - 3\sqrt x  - 3 = 0 \Leftrightarrow 4\sqrt x \left( {\sqrt x  + 1} \right) - 3\left( {\sqrt x  + 1} \right) = 0\\ \Leftrightarrow \left( {\sqrt x  + 1} \right)\left( {4\sqrt x  - 3} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\sqrt x  =  - 1\left( {ktm} \right)\\\sqrt x  = \dfrac{3}{4} \Leftrightarrow x = \dfrac{9}{{16}}\left( {tm} \right)\end{array} \right.\end{array}$

Với $x = \dfrac{9}{{16}}$ thì $P =  - 1.$

Điều kiện: $x > 0,x \ne 4,x \ne 9$

$\begin{array}{l}P = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{4 - x}}} \right):\left( {\dfrac{{\sqrt x  - 1}}{{x - 2\sqrt x }} - \dfrac{2}{{\sqrt x }}} \right)\\ = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}} \right):\left( {\dfrac{{\sqrt x  - 1}}{{\sqrt x \left( {\sqrt x  - 2} \right)}} - \dfrac{2}{{\sqrt x }}} \right)\\ = \dfrac{{4\sqrt x \left( {2 - \sqrt x } \right) + 8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{\sqrt x  - 1 - 2\left( {\sqrt x  - 2} \right)}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\ = \dfrac{{8\sqrt x  + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 2} \right)}}{{3 - \sqrt x }}\\ = \dfrac{{4\sqrt x \left( {2 + \sqrt x } \right)}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x  - 3}}\\ = \dfrac{{4x}}{{\sqrt x  - 3}}\end{array}$

Vậy $P = \dfrac{{4x}}{{\sqrt x  - 3}}$ với $x > 0,x \ne 4,x \ne 9$

Điều kiện: $x > 0,x \ne 4,x \ne 9$

$\begin{array}{l}P = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{4 - x}}} \right):\left( {\dfrac{{\sqrt x  - 1}}{{x - 2\sqrt x }} - \dfrac{2}{{\sqrt x }}} \right)\\ = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}} \right):\left( {\dfrac{{\sqrt x  - 1}}{{\sqrt x \left( {\sqrt x  - 2} \right)}} - \dfrac{2}{{\sqrt x }}} \right)\\ = \dfrac{{4\sqrt x \left( {2 - \sqrt x } \right) + 8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{\sqrt x  - 1 - 2\left( {\sqrt x  - 2} \right)}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\ = \dfrac{{8\sqrt x  + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 2} \right)}}{{3 - \sqrt x }}\\ = \dfrac{{4\sqrt x \left( {2 + \sqrt x } \right)}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x  - 3}}\\ = \dfrac{{4x}}{{\sqrt x  - 3}}\end{array}$

Vậy $P = \dfrac{{4x}}{{\sqrt x  - 3}}$ với $x > 0,x \ne 4,x \ne 9$