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

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

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 ) 

\(\dfrac{x}{\sqrt{x}-1}\ge0\) (ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\))

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\\sqrt{x}-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge0\\x\ge1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< 0\left(ktm\right)\\x\ge1\end{matrix}\right.\) (mà \(x\ne1\))

\(\Leftrightarrow x>1\)

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

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

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

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}

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(PT\Leftrightarrow2022x^2+2022x-2021x-2021=0\)

\(\Leftrightarrow2022x\left(x+1\right)-2021\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2022x-2021\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2022x-2021=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2021}{2022}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{2021}{2022}\right\}\)

giúp với ạ

\(2022x^2+x-2021=0\)

\(\Leftrightarrow2022x^2+2022x-2021x-2021=0\)

\(\Leftrightarrow2022x\left(x+1\right)-2021\left(x+1\right)=0\)

\(\Leftrightarrow\left(2022x-2021\right)\left(x+1\right)=0\Leftrightarrow x=\dfrac{2021}{2022};x=-1\)

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

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

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

chọn B