Question

Tim x:

9.(x-2)²-x²+4=0

Answer 1

Ta có: \(9\left(x-2\right)^2-x^2+4=0\)

\(\Leftrightarrow9\left(x-2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow9\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[9\left(x-2\right)-\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(9x-18-x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(8x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\8x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\8x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{20}{8}=\frac{5}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{2;\frac{5}{2}\right\}\)