Question

Tìm x, biết:

a. \(\left(2x-3\right)^2-4=0\)

b.\(x^2-2x=0\)

c.\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

Answer 1

a, \(\left(2x-3\right)^2-4=0\)

<=> \(\left(2x-3\right)^2=4\)

<=> \(\left[{}\begin{matrix}2x-3=2\\2x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy......

b, \(x^2-2x=0\)

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

Vậy............

c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x=-255\Leftrightarrow x=-\dfrac{255}{2}\)

Vậy.......