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

Ta có: \(\dfrac{x-25}{75}+\dfrac{x-15}{85}+\dfrac{x-5}{95}+\dfrac{x-145}{15}=0\)

\(\Leftrightarrow\dfrac{x-25}{75}-1+\dfrac{x-15}{85}-1+\dfrac{x-5}{95}-1+\dfrac{x-145}{15}+3=0\)

\(\Leftrightarrow\dfrac{x-100}{75}+\dfrac{x-100}{85}+\dfrac{x-100}{95}+\dfrac{x-100}{15}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}\right)=0\)

mà \(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}>0\)

nên x-100=0

hay x=100

Vậy: S={100}

⇔ 4X - 3304/323 = 0

⇔ X=3304/323/4

⇔ X=826/323

sửa đề : 

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

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

\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

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

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

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

\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

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

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

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

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

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

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

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

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

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

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

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

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

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}

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