a) (x + 2)(3x - 15) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
b) |x - 5| = 3x + 1
\(\Leftrightarrow\left[{}\begin{matrix}x< 5\\x\ge5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\\text{ko có x thỏa mãn}\end{matrix}\right.\)
=> x = 1
a)(x+2)(3x-15)=0
\(\Leftrightarrow\)x+2=0 hoac 3x-15=0
\(\Leftrightarrow\) x=-2 \(\Leftrightarrow\)3x=15
\(\Leftrightarrow\)x=5
vay s={-2;5}
b)|x-5|=3x+1
neu x-5\(\ge\)0 thi|x-5|=x-5
\(\Leftrightarrow\)x\(\ge\)5
ta co phuong trinh :
x-5=3x+1
\(\Leftrightarrow\)-2x=6
\(\Leftrightarrow\)x=-3(k thoa man dk x\(\ge\)5)
neu x-5<0 thi |x-5|=5-x
\(\Leftrightarrow\)x<5
ta co phuong trinh :
5-x=3x+1
\(\Leftrightarrow\)-4x=-4
\(\Leftrightarrow\)x=1(thoa man dk x<5)
vay s={1}
chuc bn hoc tot 
