a) x2 - 4x = 0
\(\Leftrightarrow\) x ( x - 4) = 0
khi x = 0 hoac x - 4 =0
\(\Leftrightarrow\) x = 4
Vay S = \(\left\{0,4\right\}\)
b) 3x ( x - 2021) - x + 2021 = 0
\(\Leftrightarrow\) 3x ( x - 2021) - ( x - 2021) = 0
\(\Leftrightarrow\) (x - 2021) (3x - 1) = 0
khi x - 2021 = 0 hoac 3x - 1 = 0
\(\Leftrightarrow\) x = 2021 \(\Leftrightarrow\) 3x = 1
\(\Leftrightarrow\) x = \(\dfrac{3}{1}=3\)
Vay S = \(\left\{2021,3\right\}\)
c) 4x ( x + 1) = 8x ( x + 1)
\(\Leftrightarrow\) 4x ( x + 1) - 8x ( x + 1) = 0
\(\Leftrightarrow\) -4x . ( x + 1) = 0
Khi -4x = 0 hoac x + 1 = 0
\(\Leftrightarrow\) x = 0 \(\Leftrightarrow\) x = -1
Vay S = \(\left\{0,-1\right\}\)
d) 2x ( x - 1) - 3 ( 1 - x) = 0
\(\Leftrightarrow\) 2x ( x - 1) + 3 ( x - 1) = 0
\(\Leftrightarrow\) ( x - 1) ( 2x + 3) = 0
khi x - 1 = 0 hoac 2x + 3 = 0
\(\Leftrightarrow\) x =1 \(\Leftrightarrow\) 2x = -3
\(\Leftrightarrow\) x = \(-\dfrac{3}{2}\)
Vay S = \(\left\{1,-\dfrac{3}{2}\right\}\)
e) 4x2 - 49 = 0
\(\Leftrightarrow\) (2x)2 - 72 = 0
\(\Leftrightarrow\) (2x - 7) (2x + 7) = 0
Khi 2x - 7 = 0 hoac 2x + 7 = 0
\(\Leftrightarrow\) 2x = 7 \(\Leftrightarrow\) 2x = -7
\(\Leftrightarrow\) x = \(\dfrac{7}{2}\) \(\Leftrightarrow\) x = \(-\dfrac{7}{2}\)
Vay S = \(\left\{\dfrac{7}{2},-\dfrac{7}{2}\right\}\)
f) cau nay ban co chep sai khong , nen minh sua lai de nhe
x2 - 12x = -36
\(\Leftrightarrow\) x2 - 12x + 36 = 0
\(\Leftrightarrow\) (x - 6)2 = 0
\(\Leftrightarrow\) x - 6 = 0
\(\Leftrightarrow\) x = 6
Chuc ban hoc tot
a) Ta có: \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy: S={0;4}
b) Ta có: \(3x\left(x-2021\right)-x+2021=0\)
\(\Leftrightarrow\left(x-2021\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2021;\dfrac{1}{3}\right\}\)
c) Ta có: \(4x\left(x+1\right)=8x\left(x+1\right)\)
\(\Leftrightarrow-4x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: S={0;-1}
d) Ta có: \(2x\left(x-1\right)-3\left(1-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{3}{2}\right\}\)
e) Ta có: \(4x^2-49=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{7}{2};-\dfrac{7}{2}\right\}\)
f) Ta có: \(x^2+36=12x\)
\(\Leftrightarrow x^2-12x+36=0\)
\(\Leftrightarrow\left(x-6\right)^2=0\)
\(\Leftrightarrow x-6=0\)
hay x=6
Vậy: S={6}
giúp mk vs ạ ai nhanh mk tick nha 
