Question

Пожалуйста, решите уравнение.
х^4-х^2/4х^2(х-1)=0

Answer 1

Ответ:

$x = - 1$

Объяснение:

$\frac{ {x}^{4} - {x}^{2} }{4 {x}^{2} (x - 1)} = 0 \: \: \: \: \: \: x≠0 \: \: \: \: \: x≠ 1 \\ \frac{ {x}^{2}( {x}^{2} - 1) }{4 {x}^{2} (x - 1) } =0 \\ \frac{ {x}^{2}(x - 1)(x + 1) }{4 {x}^{2}(x -1) } =0 \\ \frac{x + 1}{4} = 0 \\ x + 1 = 0 \\ x = - 1$

Answer 2

$\displaystyle \frac{x^4-x^2}{4x^2(x-1)}=0\quad\quad\quad \left[\begin{array}{ccc}x^2\neq0\\x-1\neq0\end{array}\right \rightarrow \quad \left[\begin{array}{ccc}x\neq0\\x\neq1\end{array}\right :\text{ODZ}\\\\\\\frac{x^2(x^2-1)}{4x^2(x-1)}=0\\\\\\\frac{(x-1)(x+1)}{4(x-1)}=0\\\\\\\frac{x+1}{4}=0\\\\\\x+1=0\\\\\\\boxed{x=-1}$