Question

tìm x

a) (x-2)²-4=0

b) (2x+1)²-2(2x²+3)=0

c) x(x²+x+1)-x²(x+1)-x+5=2

Answer 1

a) \(\left(x-2\right)^2-4=0\)

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

TH1:\(x-2=2\)⇒\(x=4\)

TH2:\(x-2=-2\)⇒\(x=0\)

Answer 2

a: Ta có: \(\left(x-2\right)^2-4=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b: Ta có: \(\left(2x+1\right)^2-2\left(2x^2+3\right)=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2-6=0\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)