Đáp án:
8) $B = -\dfrac13$
9) $B =\dfrac52$
Giải thích các bước giải:
$\quad B =\lim\limits_{x\to -1}\dfrac{\sqrt{7- 2x} + x -2}{x^2 -1}$
$\to B =\lim\limits_{x\to -1}\dfrac{(\sqrt{7- 2x} + x -2)(\sqrt{7- 2x} - x +2)}{(x-1)(x+1)(\sqrt{7- 2x} - x +2)}$
$\to B =\lim\limits_{x\to -1}\dfrac{7-2x - (x-2)^2}{(x-1)(x+1)(\sqrt{7- 2x} - x +2)}$
$\to B =\lim\limits_{x\to -1}\dfrac{-x^2 +2x + 3}{(x-1)(x+1)(\sqrt{7- 2x} - x +2)}$
$\to B =\lim\limits_{x\to -1}\dfrac{-(x+1)(x-3)}{(x-1)(x+1)(\sqrt{7- 2x} - x +2)}$
$\to B =\lim\limits_{x\to -1}\dfrac{3-x}{(x-1)(\sqrt{7- 2x} - x +2)}$
$\to B =\dfrac{3+1}{(-1-1)(\sqrt{7+2.1} + 1 +2)}$
$\to B =\dfrac{4}{-2(3 +3)}$
$\to B = -\dfrac13$
9) $B =\lim\limits_{x\to -1}\dfrac{2x+5-\sqrt{2x^2 + x + 8}}{x^2 + 3x +2}$
$\to B = \lim\limits_{x\to -1}\dfrac{(2x+5-\sqrt{2x^2 + x + 8})(2x+5+\sqrt{2x^2 + x + 8})}{(x+1)(x+2)(2x+5+\sqrt{2x^2 + x + 8})}$
$\to B = \lim\limits_{x\to -1}\dfrac{(2x+5)^2 - (2x^2 + x + 8)}{(x+1)(x+2)(2x+5+\sqrt{2x^2 + x + 8})}$
$\to B = \lim\limits_{x\to -1}\dfrac{2x^2 + 19x + 17}{(x+1)(x+2)(2x+5+\sqrt{2x^2 + x + 8})}$
$\to B =\lim\limits_{x\to -1}\dfrac{(x+1)(2x+17)}{(x+1)(x+2)(2x+5+\sqrt{2x^2 + x + 8})}$
$\to B =\lim\limits_{x\to -1}\dfrac{2x+17}{(x+2)(2x+5+\sqrt{2x^2 + x + 8})}$
$\to B =\dfrac{-2 +17}{(-1+2)(-2+5+\sqrt{2.1-1+8})}$
$\to B = \dfrac{15}{3+ 3}$
$\to B =\dfrac{5}{2}$
