Question

[tuấn nguyễn khải]

 4(x-5)-x²+5x² = 0

Answer 1

[Nguyễn Thị Thương Hoài]

4( x - 5) - x² + 5x² = 0

4(x-5) + 4x² = 0

4.( x - 5 + x²) =0

    x²  +x - 5  = 0

    ( x² + 2. \(\dfrac{1}{2}\) x + \(\dfrac{1}{4}\) ) - \(\dfrac{21}{4}\) = 0

    ( x + \(\dfrac{1}{2}\) )²  = \(\dfrac{21}{4}\)

    \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{21}{4}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{21}{4}}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{\sqrt{21}}{2}-\dfrac{1}{2}\\x=-\dfrac{\sqrt{21}}{2}-\dfrac{1}{2}\end{matrix}\right.\)

Answer 2

[Trần Minh Tiến]

\(4\left(x-5\right)-x^2+5x=0\\ 4\left(x-5\right)-\left(x^2-5x\right)=0\\ 4\left(x-5\right)-x\left(x-5\right)=0\\ \left(4-x\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}4-x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

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