\(a,\left(4x-\dfrac{1}{3}\right)^2+\dfrac{26}{25}=3\\ \Leftrightarrow\left(4x-\dfrac{1}{3}\right)^2=\dfrac{49}{25}\\ \Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{1}{3}=\dfrac{7}{5}\\4x-\dfrac{1}{3}=-\dfrac{7}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{30}\\x=-\dfrac{4}{15}\end{matrix}\right.\)

\(b,\left(2-\dfrac{x}{5}\right)^3-\dfrac{20}{9}=\dfrac{4}{27}\\ \Leftrightarrow\left(2-\dfrac{x}{5}\right)^3=\dfrac{64}{27}\\ \Leftrightarrow2-\dfrac{x}{5}=\dfrac{4}{3}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{10}{3}\)

\(c,\left|3x+\dfrac{1}{3}\right|-\dfrac{1}{2}=4\\\Leftrightarrow\left|3x+\dfrac{1}{3}\right|=\dfrac{9}{2}\\ \Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=\dfrac{9}{2}\left(3x+\dfrac{1}{3}\ge0\right)\\3x+\dfrac{1}{3}=-\dfrac{9}{2}\left(3x+\dfrac{1}{3}< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{25}{18}\left(x\ge-\dfrac{1}{9}\right)\left(N\right)\\x=-\dfrac{29}{18}\left(x< -\dfrac{1}{9}\right)\left(N\right)\end{matrix}\right. \)

\(\sqrt{x^2+25}=13\\ \Leftrightarrow x^2+25=169\\ \Leftrightarrow x^2=144\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\)

Ta có \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)

\(\Leftrightarrow C=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{3-2\sqrt{2}+16}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\\ =\dfrac{19-2\sqrt{2}}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{2-\sqrt{2}}\\ =\dfrac{\left(19-2\sqrt{2}\right)\left(2+\sqrt{2}\right)}{2}=\dfrac{34+15\sqrt{2}}{2}\)

Ta có \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)

\(\Leftrightarrow C=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{3-2\sqrt{2}+16}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\\ =\dfrac{19-2\sqrt{2}}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{2-\sqrt{2}}\\ =\dfrac{\left(19-2\sqrt{2}\right)\left(2+\sqrt{2}\right)}{2}=\dfrac{34+15\sqrt{2}}{2}\)

\(a,x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\\ b,2x^3-8x=2x\left(x^2-4\right)=2x\left(x-2\right)\left(x+2\right)\\ c,2x^2+6x=2x\left(x+3\right)\\ d,10x+15y=5\left(2x+3y\right)\)

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\(\left|1-7x\right|=2-x\\ \Leftrightarrow\left[{}\begin{matrix}1-7x=2-x\left(1-7x\ge0\right)\\1-7x=x-2\left(1-7x< 0\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}6x=-1\left(x\le\dfrac{1}{7}\right)\\8x=3\left(x>\dfrac{1}{7}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\left(x\le\dfrac{1}{7}\right)\left(N\right)\\x=\dfrac{3}{8}\left(x>\dfrac{1}{7}\right)\left(N\right)\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{6};\dfrac{3}{8}\right\}\)

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\(\dfrac{1}{a-b}\cdot\sqrt{3^2\left(a-b\right)^2}\left(a>b\right)\\ =\dfrac{1}{a-b}\cdot3\left|a-b\right|=\dfrac{3\left(a-b\right)}{a-b}=3\)

\(x^2-3x-5=0\\ \Delta=9+20=29\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{29}}{2}\\x=\dfrac{3+\sqrt{29}}{2}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{3+\sqrt{29}}{2};\dfrac{3-\sqrt{29}}{2}\right\}\)

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

\(x+\sqrt{9-x^2}-x\sqrt{9-x^2}=3\left(-3\le x\le3\right)\)

\(\Leftrightarrow\sqrt{9-x^2}-x\sqrt{9-x^2}=3-x\\ \Leftrightarrow9-x^2+x^2\left(9-x^2\right)-2x\sqrt{\left(9-x^2\right)^2}=9-6x+x^2\\ \Leftrightarrow9+8x^2-x^4-2x\left(9-x^2\right)=x^2-6x+9\\ \Leftrightarrow-x^4+2x^3+7x^2-12x=0\\ \Leftrightarrow-x\left(x^3-2x^2-7x+12\right)=0\Leftrightarrow-x\left(x^3-3x^2+x^2-3x-4x+12\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x^2+x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=3\left(N\right)\\x^2+x-4=0\left(1\right)\end{matrix}\right.\)

 \(\Delta\left(1\right)=1-4\left(-4\right)=17>0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1-\sqrt{17}}{2}\left(N\right)\\x=\dfrac{-1+\sqrt{17}}{2}\left(N\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;3;\dfrac{-1-\sqrt{17}}{2};\dfrac{-1+\sqrt{17}}{2}\right\}\)

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