pt <=> (16x^2-24x)+(2x-3) = 0
<=> (2x-3).(8x+1) = 0
<=> 2x-3=0 hoặc 8x+1=0
<=> x=3/2 hoặc x=-1/8
Vậy ..............
Theo bài ra ta có:\(16.x^2+22x-3=0\)
\(\Rightarrow\left(16x+22\right)x=3=1.3=3.1=\left(-1\right).\left(-3\right)=\left(-3\right).\left(-1\right)\)
Tự kẻ bẳng nha
\(16x^2+22x-3=0\)
\(\Leftrightarrow16x\left(x-\frac{1}{8}\right)+24\left(x-\frac{1}{8}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{8}\right)\left(16x+24\right)=0\)
\(\Leftrightarrow8\left(x-\frac{1}{8}\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{8}=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{8}\\x=-\frac{3}{2}\end{cases}}}\)
