Question

ПОМОГИТЕ никак не могу решить.

Answer 1

Ответ:

Задание 2.

а) $\boxed{x_{1} = 3;x_{2} = 1}$

б) $\boxed{ x_{1,2} = б\sqrt{2} ; x_{3,4} = б1 }$

Задание 3. $\boxed{x = -2}$

Задание 4. $\boxed{x_{1} = 0;x_{2} = 1}$

Объяснение:

Задание 2.

а) $\dfrac{3 - 2x}{x - 2} = -x$

ОДЗ: $x - 2\neq 0 \Longrightarrow x \neq 2$

$\dfrac{3 - 2x}{x - 2} = -x| \cdot(x -2)$

$3 - 2x = -x(x - 2)| \cdot(-1)$

$2x - 3 = x(x - 2)$

$2x - 3 = x^{2} - 2x$

$x^{2} -4x + 3= 0$

$D = 16 - 4 \cdot 1 \cdot 3 = 16 - 12 = 4 = 2^{2}$

$x_{1} = \dfrac{4 + 2}{2} = \dfrac{6}{2} = 3$

$x_{2} = \dfrac{4 - 2}{2} = \dfrac{2}{2} = 1$

б) $x^{4} - 3x^{2} + 2 = 0$

Замена: $x^{2} = t; t \geq 0$

$t^{2} - 3t +2 = 0$

$D = 9 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1 = 1^{2}$

$t_{1} = \dfrac{3 + 1}{2} = \dfrac{4}{2} = 2$

$t_{2} = \dfrac{3 - 1}{2} = \dfrac{2}{2} = 1$

$x^{2} = t_{1};x^{2} = t_{2}$

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

$\sqrt{x^{2}} = \sqrt{2} ;\sqrt{x^{2} } = \sqrt{1}$

$|x| = \sqrt{2};|x| =1$

$x_{1,2} = б\sqrt{2} ; x_{3,4} = б1$

Задание 3.

$\sqrt{2x^{2} - 4} = -x$

ОДЗ: $\displaystyle \left \{ {{2x^{2} - 4 \geq 0|:2} \atop {-x\geq 0| \cdot(-1)}} \right$  $\displaystyle \left \{ {{x^{2} - 2 \geq 0} \atop {x \leq 0}} \right$  $\displaystyle \left \{ {{(x - \sqrt{2})(x + \sqrt{2}) \geq 0} \atop {x \leq 0}} \right$

$x \in (-\infty; -\sqrt{2}]$

$(\sqrt{2x^{2} - 4})^{2} = (-x)^{2}$

$2x^{2} - 4 = x^{2}$

$x^{2} = 4$

$\sqrt{x^{2} } = \sqrt{4}$

$|x| = 2$

$x_{1} = 2$ - не входит в ОДЗ

$x_{2} = -2 \Longrightarrow \boxed{x = -2}$

Задание 4.

$\dfrac{x^{2} - x}{x^{2} - 6x + 9} - \dfrac{1}{3} = \dfrac{3 - x}{3x - 9}$

$\dfrac{x^{2} - x}{x^{2} - 6x + 9} - \dfrac{3 - x}{3x - 9} = \dfrac{1}{3}$

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

ОДЗ: $x - 3 \neq 0 \Longrightarrow x \neq 3$

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

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

$\dfrac{x^{2} - x}{(x - 3)^{2}} = 0 \Longleftrightarrow x^{2} - x = 0$

$x^{2} - x = 0$

$x(x - 1) = 0$

$x_{1} = 0 ;x_{2} = 1$