Question

Answer 1

b)

\(4x\left(x-2019\right)-x+2019=0\Rightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\)

\(\Rightarrow\left(x-2019\right)\left(4x-1\right)=0\Rightarrow\left[{}\begin{matrix}x-2019=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)

c)

\(\left(x-4\right)^2-36=0\Rightarrow\left(x-4\right)^2-6^2=0\)

\(\Rightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Rightarrow\left[{}\begin{matrix}x-4-6=0\\x-4+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

d)

\(x^2+8x+16=0\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)

e)

\(x\left(x+6\right)-7x-42=0\Rightarrow x\left(x+6\right)-7\left(x+6\right)=0\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f)

\(25x^2-9=0\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

Answer 2

ý a k ra đc đâu nhe \(x^2=-1\) thế này thì chịu