a)
\(x^2-5x+4x-20=0.\)
\(x^2-x-20=0\)
\(\left(x^2-x+\frac{1}{4}\right)-20-\frac{1}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2-\left(\frac{20.4+1}{4}\right)=0\)
\(\hept{\begin{cases}x-\frac{1}{2}-\left(\frac{20.4+1}{4}\right)=0\\x-\frac{1}{2}+\left(\frac{20.4+1}{4}\right)=0\end{cases}}\)
b) \(x^2+6x-7x-42=0\)
\(x^2-x-42=0\)
\(x^2-x+\frac{1}{4}-42-\frac{1}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2-\left(\frac{42.4+1}{4}\right)=0\) " tương tự con A
\(x^3-16x=0\)
\(x\left(x^2-16\right)=0\)
\(x=0,+4,-4\)
\(x^3-16x=0\)
\(x.\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=16\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)
Vậy \(x=0\)hoặc \(x=\pm4\)
Tham khảo nhé~
