a/ 3(x+1) - 15 = 0
<=> 3(x+1) = 15
<=> x+1 = 5
<=> x= 4
b/ \(x^2-6x+5=0\)
\(\Leftrightarrow x^2-x-5x+5=0\)
\(\Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy...
\(3\left(x+1\right)-15=0\\ \Leftrightarrow3\left(x+1\right)=15\\ \Leftrightarrow x+1=5\\ \Leftrightarrow x=4\\ x^2-6x+5=0\\ \Leftrightarrow\left(x^2-x\right)-\left(5x-5\right)=0\\ \Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
a,\(3\left(x+1\right)-15=0\)
\(3\left(x+1\right)-3.5=0\)
\(3\left(x+1-5\right)=0\)
Vì \(3\ne0\) nên trường hợp 1 loại.
\(x+1-5=0\)
\(x=4\)
b,\(x^2-6x+5=0\)
\(x^2-5x-x+5=0\)
\(\left(x^2-x\right)+\left(-5x+5\right)=0\)
\(x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy...
