Question

Tìm x biết

81x^3 - 4x = 0

Answer 1

81x³ - 4x = 0

=> x.(81x² - 4) = 0

=> \(\orbr{\begin{cases}x=0\\81x^2-4=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\81x^2=4\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x^2=\frac{4}{81}\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x\in\left\{\frac{2}{9};-\frac{2}{9}\right\}\end{cases}}\)

Answer 2

81x^3 - 4x = 0

<=>x.(81x²-4)=0

<=>x.(9x-2)(9x+2)=0

<=>x=0 hoặc x=2/9 hoặc x=-2/9

Answer 3

\(81x^3-4x=0\)

\(\Rightarrow x\left(81x^2-4\right)=0\)

\(\Rightarrow x\left(9x-2\right)\left(9x+2\right)=0\)

TH1:  \(x=0\)

TH2: \(9x-2=0\Rightarrow9x=2\Rightarrow x=\frac{2}{9}\)

TH3: \(9x+2=0\Rightarrow9x=-2\Rightarrow x=-\frac{2}{9}\)

Vậy: \(x=0\) hoặc \(x=\frac{2}{9}\) hoặc \(x=-\frac{2}{9}\)