\( \left(5x-2\right)\left(x-3\right)=0\)

\(\left[{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\)

\(\left(2x-1\right)^2+\left(x-3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{2};\dfrac{4}{3}\right\}\)

\(\left\{{}\begin{matrix}2x=0\\x^2-4=0\end{matrix}\right.\) ==>\(\left\{{}\begin{matrix}x=0\\x=+,-2\end{matrix}\right.\)

\(4x^2+4x+1+4x+2-2x^2-x\le0\)

\(\Leftrightarrow2x^2+7x+3\le0\Leftrightarrow\left(2x+1\right)\left(x+3\right)\le0\)

TH1 : \(\left\{{}\begin{matrix}2x+1\ge0\\x+3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\le-3\end{matrix}\right.\)<=> -1/2 =< x =< -3 

TH2 : \(\left\{{}\begin{matrix}2x+1\le0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\x\ge-3\end{matrix}\right.\)( vô lí ) 

\(S=\sqrt{x^2-2x+1+9}=\sqrt{\left(x-1\right)^2+9}\ge\sqrt{9}=3\)

chọn B

a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)

b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)

d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)

a) Ta có: 4x-20=0

$\iff 4x=20$

hay x=5

Vậy: S={5}

b) Ta có: $2x+x+12=0$

$\iff 3x+12=0$

$\iff 3x=-12$

hay x=-4

\(5x^2-3=0\Leftrightarrow x^2=\dfrac{3}{5}\Leftrightarrow x=\pm\sqrt{\dfrac{3}{5}}=\pm\dfrac{\sqrt{15}}{5}\)

\(4x^3+x=0\Leftrightarrow x\left(4x^2+1\right)=0\Leftrightarrow x=0;4x^2+1>0\)

\(5x^2-3=0\\ \Leftrightarrow5x^2=3\\ \Leftrightarrow x^2=\dfrac{3}{5}\\\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{3}{5}}\\x=-\sqrt{\dfrac{3}{5}}\end{matrix}\right. \)

vậy \(x=\sqrt{\dfrac{3}{5}}\) ;\(x=-\sqrt{\dfrac{3}{5}}\)

\(4x^3+x=0\\ \Leftrightarrow x\left(4x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{-1}{4}\left(vl\right)\end{matrix}\right.\)

vậy x=0

       \(5x^2-3=0 \)

<=> 5x² =3

<=> x²= \(\dfrac{3}{5}\)

<=>\(x=\sqrt{\dfrac{3}{5}}\)hay \(x=-\sqrt{\dfrac{3}{5}}\)

Vậy S={\(\sqrt{\dfrac{3}{5}}\);\(-\sqrt{\dfrac{3}{5}} \)}

     4x³+x=0

<=> x(4x²+1)=0

<=>x=0 hay 4x²+1=0

<=> x=0 hay 4x²=-1(vô lý)

Vậy S={0}

Ta có: \(x^3-5x^2+6x=0\)

\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)

Vậy: S={0;2;3}

\(x^3-5x^2+6x=0\)

\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)

\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2-3x\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)

\(S=\left\{0,2,3\right\}\)

x^3 -5x^2 + 6x=0

\(< =>x\left(x^2-5x+6\right)=0\)

\(< =>x=0;x^2-5x+6=0\)

+/ \(x^2-5x+6=0\)

\(< =>x^2-2x-3x+6=0\)

\(< =>\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(< =>x\left(x-2\right)-3\left(x-2\right)=0\)

\(< =>\left(x-2\right)\left(x-3\right)=0\)

=> \(x-2=0;x-3=0\)

=> \(x=2;3\)

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