Bài 1 :
\(A=x^2-7x+6\)
\(=x^2-x-6x+6\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x-6\right)\)
Bài 2 :
Câu a :
\(\dfrac{3}{4\left(x-5\right)}+\dfrac{7}{6x+30}=\dfrac{15}{2x^2-50}\)
ĐKXĐ : \(x\ne\pm5\)
\(\Leftrightarrow\) \(\dfrac{3}{4\left(x-3\right)}+\dfrac{7}{6\left(x+5\right)}=\dfrac{15}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\) \(\dfrac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\dfrac{14\left(x-5\right)}{12\left(x-5\right)\left(x+5\right)}=\dfrac{90}{12\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow9x+45+14x-60=90\)
\(\Leftrightarrow23x=115\)
\(\Leftrightarrow x=5\) ( Loại )
Vậy \(S=\left\{\varnothing\right\}\)
Câu b :
\(\left|2x+1\right|-5x=3\)
\(\Leftrightarrow\left|2x+1\right|=3+5x\)
Với \(2x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{2}\)
\(\Leftrightarrow2x+1=3+5x\)
\(\Leftrightarrow-3x=2\)
\(\Leftrightarrow x=-\dfrac{2}{3}\) ( Loại )
Với \(2x+1< 0\Leftrightarrow x< -\dfrac{1}{2}\)
\(\Leftrightarrow-2x-1=3+5x\)
\(\Leftrightarrow-7x=4\)
\(\Leftrightarrow x=-\dfrac{4}{7}\) ( Nhận )
Vậy \(S=\left\{-\dfrac{4}{7}\right\}\)
