$\lim_{x \to 1} \frac{\sqrt{x+24}-5}{x-1}$ 02/09/2021 Bởi Gianna $\lim_{x \to 1} \frac{\sqrt{x+24}-5}{x-1}$
Đáp án: `1/(10)` Giải thích các bước giải: `lim_{x->1} \frac{\sqrt{x+24}-5}{x-1}` `= lim_{x->1} \frac{x+24-25}{(x-1)(\sqrt{x+24}+5)}` `= lim_{x->1} \frac{x-1}{(x-1)(\sqrt{x+24}+5)}` `= lim_{x->1}\frac{1}{\sqrt{x+24}+5}` `= \frac{1}{\sqrt{1+24}+5}` `= 1/(10)` Bình luận
$\lim\limits_{x\to 1} \dfrac{\sqrt{x+24}-5}{x-1}$ $=\lim\limits_{x\to 1} \dfrac{\big(\sqrt{x+24}-5\big)\big(\sqrt{x+24}+5\big)}{(x-1)\big(\sqrt{x+24}+5\big)}$ $=\lim\limits_{x\to 1} \dfrac{x-1}{(x-1)\big(\sqrt{x+24}+5\big)}$ $=\lim\limits_{x\to 1} \dfrac1{\sqrt{x+24}+5}$ $=\dfrac1{\sqrt{1+24}+5}=\dfrac1{10}$ Bình luận
Đáp án: `1/(10)`
Giải thích các bước giải:
`lim_{x->1} \frac{\sqrt{x+24}-5}{x-1}`
`= lim_{x->1} \frac{x+24-25}{(x-1)(\sqrt{x+24}+5)}`
`= lim_{x->1} \frac{x-1}{(x-1)(\sqrt{x+24}+5)}`
`= lim_{x->1}\frac{1}{\sqrt{x+24}+5}`
`= \frac{1}{\sqrt{1+24}+5}`
`= 1/(10)`
$\lim\limits_{x\to 1} \dfrac{\sqrt{x+24}-5}{x-1}$
$=\lim\limits_{x\to 1} \dfrac{\big(\sqrt{x+24}-5\big)\big(\sqrt{x+24}+5\big)}{(x-1)\big(\sqrt{x+24}+5\big)}$
$=\lim\limits_{x\to 1} \dfrac{x-1}{(x-1)\big(\sqrt{x+24}+5\big)}$
$=\lim\limits_{x\to 1} \dfrac1{\sqrt{x+24}+5}$
$=\dfrac1{\sqrt{1+24}+5}=\dfrac1{10}$