Question

​Tìm nghiêm của đa thức sau

a ) ( 4x - 8 ) (    \(\frac{1}{2}\) -  x​ )
b ) 2x^2​  - 32

Answer 1

a) \(\left(4x-8\right)\left(\frac{1}{2}-x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}4x-8=0\\\frac{1}{2}-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{2}\end{array}\right.\)

b) \(2x^2-32=0\)

\(\Leftrightarrow2\left(x^2-16\right)=0\)

\(\Leftrightarrow x^2-16=0\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-4\end{array}\right.\)

Answer 2

\(a,\left(4x-8\right)\left(\frac{1}{2}-x\right)=0\)
\(\Leftrightarrow4\left(x-2\right)\left(\frac{1}{2}-x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\\frac{1}{2}-x=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{2}\end{array}\right.\)
\(b,2x^2-32=0\)
\(\Leftrightarrow2\left(x^2-16\right)=0\)
\(\Leftrightarrow x^2-16=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x+4=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-4\end{array}\right.\)