Question

x³- 0,25 x =0

Tìm x

Answer 1

\(x^3-0,25x=0\)

\(\Leftrightarrow x\left(x^2-0,25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\Leftrightarrow x^2=0,25\Leftrightarrow\left[{}\begin{matrix}x=0,5\\x=-0,5\end{matrix}\right.\end{matrix}\right.\)

Vậy pt có 3 nghiệm là.....

Answer 2

Ta có:

\(x^3-0,25x=0\)

\(\Rightarrow x.\left(x^2-0,25\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x^2=0,25\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)

Vậy \(x\in\left\{0;0,5;-0,5\right\}\)

Answer 3

x^3-0,25x=0

<=>x(x^2-0,25)=0

<=>x(x-0,5)(x+0,25)=0

<=>\(\left[{}\begin{matrix}x=0\\x-0,25=0\\x+0,25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,25\\x=-0,25\end{matrix}\right.\)

vậy x\(\in\left(0;\pm0,25\right)\)

Answer 4

Cách khác nhé:

\(x^3-0,25x=0\)

\(\Leftrightarrow x\left(x^2-0,25\right)=0\)

\(\Leftrightarrow x\left(x+0,5\right)\left(x-0,5\right)=0\)

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

Vậy.............