√+5−4x−−−−−√−2x2x2−1
√+5−4x−−−−−√−2x2x2−1
tìm giới hạn A= limx→12x−1−−−−−√+5−4x−−−−−√−2x2x2−1
Ta có $\dfrac{(\sqrt{2x-1}-1) +(\sqrt{5-4x}-1) -2(x^2-1)}{x^2-1}= \dfrac{(\sqrt{2x-1}-1)}{x^2-1} + \dfrac{(\sqrt{5-4x}-1)}{x^2-1} -2$
$=\dfrac{2(x-1)}{(x-1)(x+1)(\sqrt{2x-1}+1)} -\dfrac{4(x-1)}{(x-1)(x+1)(\sqrt{5-4x}+1)}-2$
$=\dfrac{2}{(x+1)(\sqrt{2x-1}+1)} -\dfrac{4}{(x+1)(\sqrt{5-4x}+1)}-2 =A$
Vậy $\lim \limits_{x\to 1} A = \dfrac{1}{2}-1-2=-\dfrac{5}{2}$
√+5−4x−−−−−√−2x2x2−1
√+5−4x−−−−−√−2x2x2−1