$\mathop {\lim }\limits_{x \to 1}\frac{x^n-nx+n-1}{x^2-2x+1}$
$\mathop {\lim }\limits_{x \to 1}\frac{x^n-1-n(x-1)}{(x-1)^2} = $
$\mathop {\lim }\limits_{x \to 1}\frac{(x-1)(x^{n-1}+x^{n-2}+...+x+1-n)}{(x-1)^2} = $
$\mathop {\lim }\limits_{x \to 1}\frac{x^{n-1}+x^{n-2}+...+x+1-n}{(x-1)} = $
$\mathop {\lim }\limits_{x \to 1}\frac{x^{n-1}-1+x^{n-2}-1+..+x-1}{(x-1)} = $
$\mathop {\lim }\limits_{x \to 1}\frac{x^{n-1}-1}{x-1}+\mathop {\lim }\limits_{x \to 1}\frac{x^{n-2}-1}{x-1}+...+\mathop {\lim }\limits_{x \to 1}\frac{x-1}{x-1}=$
$\mathop {\lim }\limits_{x \to 1}(x^{n-2}+x^{n-3}+....+x+1)+\mathop {\lim }\limits_{x \to 1}(x^{n-3}+x^{n-4}+....+x+1)+...+\mathop {\lim }\limits_{x \to 1}(1)=$
$(n-1)+(n-2)+...+2+1  = \frac{n(n-1)}{2}$

Làm đúng nhớ vote nhé, ủng hộ tinh thần

$\mathop {\lim }\limits_{x \to 1}\frac{x^n-1-n(x-1)}{(x-1)^2}=\frac{x^{n-1}+x^{n-2}+...+x-n}{(x-1)}$
$=\frac{n(x-1)}{x-1}=n$
uk thieu so 1 –  Nero 12-05-14 02:01 PM
Kiểm tra lại nhé, có sự nhầm lẫn, có thể dung L'hopitan để kiểm tra lại kết quả. –  diendien_01 11-05-14 02:08 AM
thấy đúng thì vote cho đáp án nhé! Lần sau mình sẽ ss giúp đỡ. Tks bạn! –  Nero 10-05-14 01:01 PM

Bạn cần đăng nhập để có thể gửi đáp án

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