Giải:
\(\left(2x-1\right)^2-\left(2x-3\right)\left(2x+3\right)=8\)
\(\Leftrightarrow4x^2-4x+1-\left(4x^2-3^2\right)=8\)
\(\Leftrightarrow4x^2-4x+1-4x^2+3^2=8\)
\(\Leftrightarrow-4x+1+3^2=8\)
\(\Leftrightarrow-4x+10=8\)
\(\Leftrightarrow-4x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{-4}\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\).
Chúc bạn học tốt!!!
\(\left(2x-1\right)^2-\left(2x-3\right)\left(2x+3\right)=8\)
\(\Leftrightarrow4x^2-4x+1-4x^2+9=8\)
\(\Leftrightarrow-4x=-2\Leftrightarrow x=\dfrac{1}{2}\)
Vậy...
Giải.
Ta có :\(\left(2x-1\right)^2-\left(2x-3\right)\left(2x+3\right)=8\)
\(\Leftrightarrow\left(2x\right)^2-2x+1-\left[\left(2x^2\right)-9\right]=8\)
\(\Leftrightarrow\left(2x\right)^2-2x+1-\left(2x\right)^2-9=8\)
\(\Leftrightarrow-2x-8=8\)
\(\Leftrightarrow-2x=16\Leftrightarrow x=-8\)
Vậy x = -8
