Cho C = lim [{x^2-mx+m-1} / {x^2-1}], m là số thực. x–>1 Tìm m để C=2 25/10/2021 Bởi Ximena Cho C = lim [{x^2-mx+m-1} / {x^2-1}], m là số thực. x–>1 Tìm m để C=2
`C=lim_{x->1}(x^2-mx+m-1)/(x^2-1)` `=lim_{x->1}((x-1)(x+1)-m(x-1))/((x-1)(x+1))` `=lim_{x->1}((x-1)(x+1-m))/((x-1)(x+1))` `=lim_{x->1}(x+1-m)/(x+1)` `=(2-m)/3` Ta có: `C=2` `<=>(2-m)/3=2` `<=>2-m=6` `<=>m=-4` Vậy `m=-4` thì `C=2` Bình luận
$\lim\limits_{x\to 1}\dfrac{(x^2-mx+m-1}{(x-1)(x+1)}$ $=\lim\limits_{x\to 1}\dfrac{x^2-1-mx+m}{(x-1)(x+1)}$ $=\lim\limits_{x\to 1}\dfrac{(x-1)(x+1)-m(x-1) }{(x-1)(x+1)}$ $=\lim\limits_{x\to 1}\dfrac{x+1-m}{x+1}$ $=\dfrac{2-m}{2}=2$ $\to m=-2$ Bình luận
`C=lim_{x->1}(x^2-mx+m-1)/(x^2-1)`
`=lim_{x->1}((x-1)(x+1)-m(x-1))/((x-1)(x+1))`
`=lim_{x->1}((x-1)(x+1-m))/((x-1)(x+1))`
`=lim_{x->1}(x+1-m)/(x+1)`
`=(2-m)/3`
Ta có:
`C=2`
`<=>(2-m)/3=2`
`<=>2-m=6`
`<=>m=-4`
Vậy `m=-4` thì `C=2`
$\lim\limits_{x\to 1}\dfrac{(x^2-mx+m-1}{(x-1)(x+1)}$
$=\lim\limits_{x\to 1}\dfrac{x^2-1-mx+m}{(x-1)(x+1)}$
$=\lim\limits_{x\to 1}\dfrac{(x-1)(x+1)-m(x-1) }{(x-1)(x+1)}$
$=\lim\limits_{x\to 1}\dfrac{x+1-m}{x+1}$
$=\dfrac{2-m}{2}=2$
$\to m=-2$