Đưa thừa số ra ngoài dấu căn và đưa thừa số vào trong dấu căn

Ta có

$ \begin{array}{l} \sqrt{16\left( {{x}^{2}}-1 \right)}=12 \\ \Leftrightarrow 16\left( {{x}^{2}}-1 \right)={{12}^{2}} \\ \Leftrightarrow {{x}^{2}}-1=\dfrac{{{12}^{2}}}{16}=9 \\ \Leftrightarrow {{x}^{2}}=10 \\ \Leftrightarrow x=\pm \sqrt{10} \end{array} $

Vậy tổng bình phương các giá trị của x thỏa mãn bằng: $ {{\left( -\sqrt{10} \right)}^{2}}+{{\left( \sqrt{10} \right)}^{2}}=10+10=20 $

Ta có: $ x\sqrt{\dfrac{-35}{x}} $ $ =-\sqrt{{{x}^{2}}.\dfrac{-35}{x}}=-\sqrt{-35x} $

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

$ =\left( \dfrac{\sqrt{7}\left( \sqrt{2}-1 \right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left( \sqrt{3}-1 \right)}{1-\sqrt{3}} \right).\left( \sqrt{7}-\sqrt{5} \right) $

$ =\left( -\sqrt{7}-\sqrt{5} \right).\left( \sqrt{7}-\sqrt{5} \right)=-\left( \sqrt{7}+\sqrt{5} \right)\left( \sqrt{7}-\sqrt{5} \right)=-\left( 7-5 \right)=-2 $

Ta có

$\begin{gathered} 5\sqrt a - 4b\sqrt {25{a^3}} + 5a\sqrt {16a{b^2}} - \sqrt {9a} \hfill \\ = 5\sqrt a - 4\sqrt {25{a^3}{b^2}} + 5\sqrt {16a{b^2}.{a^2}} - \sqrt 9 .\sqrt a \hfill \\ \end{gathered}$

$ =5\sqrt{a}-4\sqrt{25}.\sqrt{{{a}^{3}}{{b}^{2}}}+5\sqrt{16}.\sqrt{{{a}^{3}}{{b}^{2}}}-3\sqrt{a} $ $ =\left( 5\sqrt{a}-3\sqrt{a} \right)-\left( 4.5\sqrt{{{a}^{3}}{{b}^{2}}}-5.4\sqrt{{{a}^{3}}{{b}^{3}}} \right) $ $ =2\sqrt{a} $ .

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

$ \begin{array}{l} A=5\sqrt{12}+2\sqrt{75}-5\sqrt{48} \\ =5\sqrt{3.4}+2\sqrt{25.3}-5\sqrt{16.3} \\ =10\sqrt{3}+10\sqrt{3}-20\sqrt{3} \\ =0 \end{array} $

Vì $ x < 0;y > 0 $ nên ta có

$ \begin{array}{l} -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}}}}{(-{{x}^{3}}{{y}^{2}})} \\ =2.\dfrac{\sqrt{-{{3}^{2}}x}.\left| x \right|.\left| y \right|}{xy}=\dfrac{2.3\sqrt{-x}(-x).y}{xy}=-6\sqrt{-x}. \end{array} $

$ 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 $ .

ĐKXĐ: ${x^2}\left( {x - 1} \right) \ge 0 \Leftrightarrow x = 0;x \ge 1$

Khi đó ta có

TH1: x = 0 là nghiệm của phương trình.

TH2: \[x \ne 0;x \ge 1\]

$\begin{array}{*{20}{l}} {Pt \Leftrightarrow \left| x \right|\sqrt {x - 1} = 2x}\\ { \Leftrightarrow x\sqrt {x - 1} - 2x = 0}\\ { \Leftrightarrow x\left( {\sqrt {x - 1} - 2} \right) = 0}\\ { \Leftrightarrow \left[ {\begin{array}{*{20}{l}} {x = 0(L)}\\ {\sqrt {x - 1} = 2} \end{array}} \right. \Leftrightarrow x - 1 = 4 \Leftrightarrow x = 5} \end{array}$

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}-\dfrac{4}{2} \right)=\dfrac{\sqrt{6}}{6} $

Ta có $ \sqrt{32x}+\sqrt{50x}-2\sqrt{8x}+\sqrt{18x} $

$ =\sqrt{16.2x}+\sqrt{25.2x}-2\sqrt{4.2x}+\sqrt{9.2x}=\sqrt{{{4}^{2}}.2x}+\sqrt{{{5}^{2}}.2x}-2\sqrt{{{2}^{2}}.2x}+\sqrt{{{3}^{2}}.2x} $ $ =4\sqrt{2x}+5\sqrt{2x}-4\sqrt{2x}+3\sqrt{2x}=\sqrt{2x}(4+5-4+3)=8\sqrt{2x} $ .

$ \begin{array}{l} A=3\sqrt{8}-4\sqrt{18}+2\sqrt{50} \\ =3\sqrt{2.4}-4.\sqrt{2.9}+2.\sqrt{2.25} \\ =3.2.\sqrt{2}-4.3.\sqrt{2}+2.5.\sqrt{2} \\ =6\sqrt{2}-12\sqrt{2}+10\sqrt{2} \\ =4\sqrt{2} \end{array} $

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

Ta có $ \sqrt{8}=2\sqrt{2};\sqrt{18}=3\sqrt{2} $ , khi đó

$ \begin{array}{l} A=\left( 2\sqrt{2}-\sqrt{5}+3\sqrt{2} \right)\left( 3\sqrt{2}+\sqrt{5}+2\sqrt{2} \right) \\ A=\left( 5\sqrt{2}-\sqrt{5} \right)\left( 5\sqrt{2}+\sqrt{5} \right) \\ A={{\left( 5\sqrt{2} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}=50-5=45 \end{array} $

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

$ 2\sqrt{\dfrac{16a}{3}}-3\sqrt{\dfrac{a}{27}}-6\sqrt{\dfrac{4a}{75}}=2\sqrt{{{4}^{2}}.\dfrac{a}{3}}-3\sqrt{\dfrac{1}{9}.\dfrac{a}{3}}-6\sqrt{\dfrac{4}{25}.\dfrac{a}{3}}=2.4\sqrt{\dfrac{a}{3}}-3.\dfrac{1}{3}\sqrt{\dfrac{a}{3}}-6.\dfrac{2}{5}.\sqrt{\dfrac{a}{3}} $

$ =\sqrt{\dfrac{a}{3}}.\left( 8-1-\dfrac{12}{5} \right)=\dfrac{23}{5}\sqrt{\dfrac{a}{3}}=\dfrac{23}{5}.\dfrac{\sqrt{3a}}{3}=\dfrac{23\sqrt{3a}}{15} $ .

Ta có $ 5y\sqrt{y}=\sqrt{{{(5y)}^{2}}y}=\sqrt{25{{y}^{2}}.y}=\sqrt{25{{y}^{3}}} $ .

Rút gọn biểu thức ta được $ Q=\dfrac{\sqrt{x-y}}{\sqrt{x+y}} $ với $ x > y > 0 $

Thay $ x=3y $ (thỏa mãn ĐK) vào biểu thức $ Q $ , ta được $ Q=\dfrac{\sqrt{3y-y}}{\sqrt{3y+y}}=\dfrac{\sqrt{2y}}{\sqrt{4y}}=\dfrac{\sqrt{2}}{2} $

Vậy $ Q=\dfrac{\sqrt{2}}{2} $ khi $ x=3y $ .

$ \sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80} $

$ =\sqrt{25.5}-4\sqrt{9.5}+3\sqrt{4.5}-\sqrt{16.5}=5\sqrt{5}-4.3\sqrt{5}+3.2\sqrt{5}-4\sqrt{5} $

$ =5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}. $

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

Ta có

$ \begin{array}{l} \dfrac{1}{2ab}\sqrt{12{{a}^{3}}{{b}^{2}}{{c}^{5}}} \\ =\dfrac{1}{2ab}\sqrt{3.4.{{a}^{2}}.a.{{b}^{2}}.{{c}^{4}}.c} \\ =\dfrac{1}{2ab}.2.\left| a \right|.\left| b \right|{{c}^{2}}\sqrt{3ac} \\ =-{{c}^{2}}\sqrt{3ac} \end{array} $

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} $

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

Điều kiện: $x>-1$

Ta có $ \sqrt{{{x}^{2}}-6x+9}=x+1\Leftrightarrow \sqrt{{{\left( x-3 \right)}^{2}}}=x+1 $

$ \begin{array}{l} \Leftrightarrow \left| x-3 \right|=x+1 \\ \Leftrightarrow \left[ \begin{array}{l} x-3=x+1 \\ 3-x=x+1 \end{array} \right. \\ \Leftrightarrow x=1 \end{array} $

Ta có $ \sqrt{9x}\le 15\Leftrightarrow \left\{ \begin{array}{l} x\ge 0 \\ 9x\le {{15}^{2}} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x\ge 0 \\ x\le 25 \end{array} \right. $

Vậy có 26 giá trị nguyên thỏa mãn.

Ta có $ \sqrt{81{{(2-y)}^{4}}}=\sqrt{81.{{\left[ {{(2-y)}^{2}} \right]}^{2}}}=\sqrt{81}\left| {{(2-y)}^{2}} \right|=9{{(2-y)}^{2}} $

Điều kiện: $ \left\{ \begin{array}{l} 4x-8\ge 0 \\ 9x-18\ge 0 \\ x-24\ge 0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 4(x-2)\ge 0 \\ 9(x-2)\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 $

$ \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

$ \begin{array}{l} x+\sqrt{3}=2\Leftrightarrow 2-x=\sqrt{3}\Rightarrow {{(2-x)}^{2}}=3 \\ \Leftrightarrow 4-4x+{{x}^{2}}=3\Leftrightarrow {{x}^{2}}-4x+1=0. \end{array} $

Suy ra $ H=({{x}^{5}}-4{{x}^{4}}+{{x}^{3}})+({{x}^{4}}-4{{x}^{3}}+{{x}^{2}})+5({{x}^{2}}-4x+1)+2019. $

Do $ {{x}^{2}}-4x+1=0 $ nên $ H=2019. $