\(3x\left(x-2\right)-3\left(x^2-4\right)=0\\ =>3x\left(x-2\right)-3\left(x-2\right)\left(x+2\right)=0\\ =>\left(x-2\right)\left[3x-3\left(x+2\right)\right]=0\\ =>\left[{}\begin{matrix}x-2=0\\3x-3x-6=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=2\\-6=0\left(voli\right)\end{matrix}\right.\\ =>x=2\)

\(=8x^2-2x-12x+3-2\left(1-4x+4x^2\right)\\ =8x^2-14x+3-2+8x-8x^2\\ =-6x+1\)

\(đkx\ge0\\ 2x+5\sqrt{x}-3=0\\ =>2x-\sqrt{x}+6\sqrt{x}-3=0\\ =>\sqrt{x}\left(2\sqrt{x}-1\right)+3\left(2\sqrt{x}-1\right)=0\\ =>\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}+3=0\\2\sqrt{x}-1=0\end{matrix}\right.=>\left[{}\begin{matrix}\sqrt{x}=-3\left(voli\right)\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\\ =>x=\dfrac{1}{4}\left(thoaman\right)\)

\(\left|\overrightarrow{AB}-\overrightarrow{AD}\right|=\left|\overrightarrow{AB}+\overrightarrow{DA}\right|=\left|\overrightarrow{DB}\right|=DB=AC\\ AC=\sqrt{a^2+\left(2a\right)^2-2.a.2a.cos120^o}=\sqrt{3}a=BD\\ =>\left|\overrightarrow{AB}-\overrightarrow{AD}\right|=\sqrt{3}a\)

_{:v thuốc thông não nek} 

anh ác lắm biết cam mờ xong cứ đòi hình 

\(a,A=\left\{0;1;2;3;4;5;6\right\}\)

\(b,\)

\(20\%=\dfrac{1}{5}\\ \dfrac{1}{5}\times155=31\)

\(c,=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(5-\sqrt{3}\right)^2}\\ =\left|\sqrt{3}+2\right|+\left|5-\sqrt{3}\right|\\ =\sqrt{3}+2+5-\sqrt{5}\\ =7\)

\(d,=\dfrac{14\left(\sqrt{10}-\sqrt{3}\right)}{10-3}-2\sqrt{10}\\ =2\left(\sqrt{10}-\sqrt{3}\right)-2\sqrt{10}\\ =2\sqrt{10}-2\sqrt{3}-2\sqrt{10}\\ =-2\sqrt{3}\)