tính 4\lim\limits{x\to1} \dfrac{x^2-x-2}{x+1}$ 23/10/2021 Bởi Audrey tính 4\lim\limits{x\to1} \dfrac{x^2-x-2}{x+1}$
`~rai~` \(\lim\limits_{x\to 1} \dfrac{x^2-x-2}{x+1}\\=\lim\limits_{x\to 1}\dfrac{x^2+x-2x-2}{x+1}\\=\lim\limits_{x\to 1} \dfrac{x\left(x+1\right)-2\left(x+1\right)}{x+1}\\=\lim\limits_{x\to 1} \dfrac{\left(x-2\right)\left(x+1\right)}{x+1}\\=\lim\limits_{x\to 1} x-2\\=1-2=-1.\) Bình luận
Đáp án: `lim_(x->1)(x^2-x-2)/(x+1)=-1` Giải thích các bước giải: `lim_(x->1)(x^2-x-2)/(x+1)` `=lim_(x->1)((x-2)(x+1))/(x+1)` `=lim_(x->1)(x-2)` `=1-2=-1` Bình luận
`~rai~`
\(\lim\limits_{x\to 1} \dfrac{x^2-x-2}{x+1}\\=\lim\limits_{x\to 1}\dfrac{x^2+x-2x-2}{x+1}\\=\lim\limits_{x\to 1} \dfrac{x\left(x+1\right)-2\left(x+1\right)}{x+1}\\=\lim\limits_{x\to 1} \dfrac{\left(x-2\right)\left(x+1\right)}{x+1}\\=\lim\limits_{x\to 1} x-2\\=1-2=-1.\)
Đáp án:
`lim_(x->1)(x^2-x-2)/(x+1)=-1`
Giải thích các bước giải:
`lim_(x->1)(x^2-x-2)/(x+1)`
`=lim_(x->1)((x-2)(x+1))/(x+1)`
`=lim_(x->1)(x-2)`
`=1-2=-1`