Question

8 класс , алгебра
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Answer 1

Ответ:

Решения уравнений:

1) $\boldsymbol{\boxed{x=1}}$

2) $\boldsymbol{\boxed{ x_{1,2} = \pm 3; x_{3,4} = \pm 2}}$

Объяснение:

1)

$\dfrac{x^{2} +1}{x-3} = \dfrac{2x}{x-3}$

ОДЗ: $x \in \mathbb R \backslash \{3\}$

$\dfrac{x^{2} +1}{x-3} = \dfrac{2x}{x-3} \bigg | \cdot (x- 3)$

$x^{2} +1 = 2x$

$x^{2} - 2x + 1 = 0$

$(x - 1)^{2} = 0 \Longleftrightarrow x - 1 = 0 \Longrightarrow x = 1$

2)

$x^{4} - 13x^{2} + 36 = 0$

Замена: $x^{2} = t; t \geq 0$

$t^{2} - 13t + 36 = 0$

$D = 169 - 4 \cdot 1 \cdot 36 = 169 - 144 = 25 = 5^{2}$

$t_{1} = \dfrac{13+ 5}{2} = \dfrac{18}{2} = 9$

$t_{2} = \dfrac{13 - 5}{2} = \dfrac{8}{2} = 4$

$\left[ \begin{gathered} x^{2} = t_{1} \\ x^{2} =t_{2} \\ \end{gathered} \right.$  $\left[ \begin{gathered} x^{2} =9 \\ x^{2} =4 \\ \end{gathered} \right.$  $\left[ \begin{gathered} \sqrt{x^{2} } =\sqrt{9} \\ \sqrt{x^{2} } =\sqrt{4} \\ \end{gathered} \right.$  $\left[ \begin{gathered} |x| =3 \\ |x| =2 \\ \end{gathered} \right.$   $\left[ \begin{gathered} x_{1,2} = \pm 3 \\ x_{3,4} = \pm 2 \\ \end{gathered} \right.$