Bài 1:
a) \(x\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
b) \(\left(3-x\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-0\\x^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\sqrt{-1}\end{matrix}\right.\)
Vì một số không âm mới có căn bậc hai \(\Rightarrow x=3\)
c) \(\left(x-1\right)^2=4\)
\(\Leftrightarrow\left(x-1\right)^2=2^2=\left(-2\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+1\\x=\left(-2\right)+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
d) \(\left(2x-1\right)^3=8\)
\(\Leftrightarrow\left(2x-1\right)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
\(\Leftrightarrow2x=2+1\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\) (không thỏa mãn vì \(x\in Z\))
Vậy \(x\in\varnothing\)
e) \(\left(x+3\right)^2=16\)
\(\Leftrightarrow\left(x+3\right)^2=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4-3\\x=\left(-4\right)-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
