Question

Answer 1

a) \(2x^2-8=0\)

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

\(x^2-4=0\)

\(x^2=4\)

⇒\(x=+-2\)

Answer 2

a) \(\Rightarrow x^2-4=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

b) => 4x.(x - 2) - (x - 2) = 0

=> (x - 2).(4x - 1) = 0

=> x = 2 hoặc x = 1/4

c) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

=> (x + 3).x(x - 2) = 0

=> x = -3 hoặc x = 0 hoặc x = 2

Answer 3

b) \(4x\left(x-2\right)-x+2=0\)

\(4x\left(x-2\right)-\left(x-2\right)=0\)

\(\left(4x-1\right)\left(x-2\right)=0\)

TH1:4x=1⇒\(x=\dfrac{1}{4}\)

TH2:x-2=0⇒x=2