Question

Задание приложено...

Answer 1

Ответ:

Пределы:

1) $\boxed{ \boldsymbol{ \displaystyle \lim_{x \to -3} \bigg(\frac{1}{x + 3} + \frac{6}{x^{2} - 9} \bigg) = -\frac{1}{6} } }$

2) $\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 2} \bigg(\frac{2}{2x^{2} - 5x + 2} + \frac{x - 4}{3(x^{2} - 3x + 2)} \bigg) = \frac{5}{9} } }$

Объяснение:

38.6

1)

$\displaystyle \lim_{x \to -3} \bigg(\frac{1}{x + 3} + \frac{6}{x^{2} - 9} \bigg) = -\frac{1}{6}$

а) преобразуем выражение $\displaystyle \bigg(\frac{1}{x + 3} + \frac{6}{x^{2} - 9} \bigg) :$

$\displaystyle \bigg ( \frac{1}{x + 3} + \frac{6}{x^{2} - 9} \bigg) = \frac{1}{x + 3} + \frac{6}{x^{2} - 9} = \frac{1}{x + 3} + \frac{6}{(x - 3)(x + 3)} = \frac{x - 3+6}{(x - 3)(x + 3)}=$

$\displaystyle = \frac{x + 3}{(x - 3)(x + 3)}= \frac{1}{x -3}$

$\displaystyle \lim_{x \to -3} \bigg(\frac{1}{x + 3} + \frac{6}{x^{2} - 9} \bigg) = \lim_{x \to -3} \frac{1}{x -3} = \frac{1}{-3 -3} = -\frac{1}{6}$

2)

$\displaystyle \lim_{x \to 2} \bigg(\frac{2}{2x^{2} - 5x + 2} + \frac{x - 4}{3(x^{2} - 3x + 2)} \bigg) = \lim_{x \to 2} \frac{ (x +0,5)}{3(x -1)(x -0,5)} =$

$= \dfrac{ (2 +0,5)}{3(2 -1)(2 -0,5)} = \dfrac{2,5}{3 \cdot1,5} = \dfrac{5}{9}$

а) преобразуем выражение $\displaystyle \bigg(\frac{2}{2x^{2} - 5x + 2} + \frac{x - 4}{3(x^{2} - 3x + 2)} \bigg) :$

$\displaystyle \bigg(\frac{2}{2x^{2} - 5x + 2} + \frac{x - 4}{3(x^{2} - 3x + 2)} \bigg) = \frac{2}{2x^{2} - 5x + 2} + \frac{x - 4}{3(x^{2} - 3x + 2)} =$

$\displaystyle = \frac{2}{2(x - 2)(x -0,5)} + \frac{x - 4}{3 (x - 2)(x - 1)} = \frac{2 \cdot3(x - 1) +2(x - 0,5)(x - 4)}{6(x - 2)(x -1)(x -0,5)} =$

$\displaystyle = \frac{ 2(x - 2)(x +0,5)}{6(x - 2)(x -1)(x -0,5)} = \frac{ (x +0,5)}{3(x -1)(x -0,5)}$

б) разложим на множители выражение $2x^{2} - 5x + 2:$

$2x^{2} - 5x + 2 = 0$

$D = 25 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 = 3^{2}$

$\displaystyle x_{1} = \frac{5 + 3}{2 \cdot 2} = \frac{8}{4} = 2$

$\displaystyle x_{2} = \frac{5 - 3}{2 \cdot 2} = \frac{2}{4} = 0,5$

$2x^{2} - 5x + 2 = 2(x - 2)(x -0,5)$

в) разложим на множители выражение $2x^{2} - 5x + 2:$

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

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

$\displaystyle x_{1} = \frac{3 + 1}{2} = \frac{4}{2} = 2$

$\displaystyle x_{2} = \frac{3 - 1}{2} = \frac{2}{2} = 1$

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

г) преобразуем выражение $2 \cdot3(x - 1) +2(x - 0,5)(x - 4):$

$2 \cdot3(x - 1) +2(x - 0,5)(x - 4) = 6(x - 1) + (2x - 1)(x - 4) =$

$= 6x - 6 + 2x^{2} -8x - x + 4 = 2x^{2} + 6x - 9x - 6 + 4 = 2x^{2} -3x -2 =$

$= 2(x - 2)(x +0,5)$

Разложим на множители: $2x^{2} -3x -2$

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

$D = 9 -4 \cdot 2 \cdot (-2) = 9 + 16 = 25 = 5^{2}$

$\displaystyle x_{1} = \frac{3 + 5}{2 \cdot 2} = \frac{8}{4} = 2$

$\displaystyle x_{2} = \frac{3 - 5}{2 \cdot 2} = \frac{-2}{4} = -0,5$

$2x^{2} -3x -2 = 2(x - 2)(x +0,5)$