Решение:
![\displaystyle\mathtt{(k+1)x^2-2x+1-k=0;~[a=k+1;~b=-2;~c=1-k]~D=b^2-}\\\displaystyle\mathtt{-4ac=(-2)^2-4(k+1)(1-k)=4[1-(1-k)(1+k)]=}\\\displaystyle\mathtt{=4(1-[1^2-k^2])=4(1-1+k^2)=4k^2;}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7B%28k%2B1%29x%5E2-2x%2B1-k%3D0%3B~%5Ba%3Dk%2B1%3B~b%3D-2%3B~c%3D1-k%5D~D%3Db%5E2-%7D%5C%5C%5Cdisplaystyle%5Cmathtt%7B-4ac%3D%28-2%29%5E2-4%28k%2B1%29%281-k%29%3D4%5B1-%281-k%29%281%2Bk%29%5D%3D%7D%5C%5C%5Cdisplaystyle%5Cmathtt%7B%3D4%281-%5B1%5E2-k%5E2%5D%29%3D4%281-1%2Bk%5E2%29%3D4k%5E2%3B%7D)
первый вариант, когда
![\displaystyle\mathtt{D\ \textless \ 0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7BD%5C+%5Ctextless+%5C+0%7D)
;
ответ: решений нет; решение:
![\displaystyle\mathtt{D\ \textless \ 0~\to~4k^2\ \textless \ 0~\to~k^2\ \textless \ 0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7BD%5C+%5Ctextless+%5C+0~%5Cto~4k%5E2%5C+%5Ctextless+%5C+0~%5Cto~k%5E2%5C+%5Ctextless+%5C+0%7D)
второй вариант, когда
![\displaystyle\mathtt{D=0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7BD%3D0%7D)
;
ответ:
при ![\displaystyle\mathtt{k=0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7Bk%3D0%7D)
; решение:
![\displaystyle\mathtt{D=0~\to~4k^2=0~\to~k^2=0~\to~k=0~\to~x=\frac{-bб\sqrt{D}}{2a}=\frac{-b}{2a}=}\\\displaystyle\mathtt{=\frac{-(-2)}{2(k+1)}=\frac{2}{2(k+1)}=\frac{1}{k+1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7BD%3D0~%5Cto~4k%5E2%3D0~%5Cto~k%5E2%3D0~%5Cto~k%3D0~%5Cto~x%3D%5Cfrac%7B-b%D0%B1%5Csqrt%7BD%7D%7D%7B2a%7D%3D%5Cfrac%7B-b%7D%7B2a%7D%3D%7D%5C%5C%5Cdisplaystyle%5Cmathtt%7B%3D%5Cfrac%7B-%28-2%29%7D%7B2%28k%2B1%29%7D%3D%5Cfrac%7B2%7D%7B2%28k%2B1%29%7D%3D%5Cfrac%7B1%7D%7Bk%2B1%7D%7D)
третий вариант, когда
![\displaystyle\mathtt{D\ \textgreater \ 0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7BD%5C+%5Ctextgreater+%5C+0%7D)
; собственно, решение:
![\displaystyle\mathtt{D\ \textgreater \ 0~\to~4k^2\ \textgreater \ 0~\to~k^2\ \textgreater \ 0~\to~k\in(-\infty;0)(0;+\infty)~\to~x=}\\\displaystyle\mathtt{=\frac{-bб\sqrt{D}}{2a}=\frac{-(-2)б\sqrt{4k^2}}{2(k+1)}}=\frac{2б2|k|}{2(k+1)}=\frac{б|k|+1}{k+1}};](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7BD%5C+%5Ctextgreater+%5C+0~%5Cto~4k%5E2%5C+%5Ctextgreater+%5C+0~%5Cto~k%5E2%5C+%5Ctextgreater+%5C+0~%5Cto~k%5Cin%28-%5Cinfty%3B0%29%280%3B%2B%5Cinfty%29~%5Cto~x%3D%7D%5C%5C%5Cdisplaystyle%5Cmathtt%7B%3D%5Cfrac%7B-b%D0%B1%5Csqrt%7BD%7D%7D%7B2a%7D%3D%5Cfrac%7B-%28-2%29%D0%B1%5Csqrt%7B4k%5E2%7D%7D%7B2%28k%2B1%29%7D%7D%3D%5Cfrac%7B2%D0%B12%7Ck%7C%7D%7B2%28k%2B1%29%7D%3D%5Cfrac%7B%D0%B1%7Ck%7C%2B1%7D%7Bk%2B1%7D%7D%3B)
![\displaystyle\mathtt{x=\frac{б|k|+1}{k+1}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7Bx%3D%5Cfrac%7B%D0%B1%7Ck%7C%2B1%7D%7Bk%2B1%7D%7D%7D)
для
![\displaystyle\mathtt{k\neq0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7Bk%5Cneq0%7D)
, теперь стоит учесть ОДЗ:
![\displaystyle\mathtt{k+1\neq0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7Bk%2B1%5Cneq0%7D)
, как знаменатель, поэтому наша половина ответа выглядит так: при
![\displaystyle\mathtt{k\in(-\infty;-1)(-1;0)(0;+\infty)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7Bk%5Cin%28-%5Cinfty%3B-1%29%28-1%3B0%29%280%3B%2B%5Cinfty%29%7D)
корень уравнения ищется по формуле
![\displaystyle\mathtt{x=\frac{б|k|+1}{k+1}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7Bx%3D%5Cfrac%7B%D0%B1%7Ck%7C%2B1%7D%7Bk%2B1%7D%7D%7D)
итак, поработав немного с модулем, можно вывести окончательный
ответ: 1.
при
; 2.
при
.