![(a+1)x^2-2x+1-а=0 \\(a+1)x^2-2x+(1-a)=0 \\D=4-4(a+1)(1-a)=4(1+a^2-1)=4a^2](https://tex.z-dn.net/?f=%28a%2B1%29x%5E2-2x%2B1-%D0%B0%3D0%0A%5C%5C%28a%2B1%29x%5E2-2x%2B%281-a%29%3D0%0A%5C%5CD%3D4-4%28a%2B1%29%281-a%29%3D4%281%2Ba%5E2-1%29%3D4a%5E2)
1) при a+1=0; a=-1 уравнение обращается в линейное:
![-2x+1+1=0 \\-2x=-2 \\x=1](https://tex.z-dn.net/?f=-2x%2B1%2B1%3D0%0A%5C%5C-2x%3D-2%0A%5C%5Cx%3D1)
2) при D>0 и a≠-1 имеет 2 различных корня
![\left \{ {{4a^2\ \textgreater \ 0} \atop {a \neq -1}} \right. \Rightarrow \left \{ {{a \in (-\infty;0)\cup (0;+\infty)} \atop {a \neq -1}} \right. \Rightarrow a \in (-\infty;-1)\cup (-1;0)\cup (0;+\infty) \\x_1= \frac{2+\sqrt{4a^2}}{2(a+1)} = \frac{1+|a|}{a+1} \\x_2= \frac{1-|a|}{a+1}](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B4a%5E2%5C+%5Ctextgreater+%5C+0%7D+%5Catop+%7Ba+%5Cneq+-1%7D%7D+%5Cright.+%5CRightarrow++%5Cleft+%5C%7B+%7B%7Ba+%5Cin+%28-%5Cinfty%3B0%29%5Ccup+%280%3B%2B%5Cinfty%29%7D+%5Catop+%7Ba+%5Cneq+-1%7D%7D+%5Cright.++%5CRightarrow+a+%5Cin+%28-%5Cinfty%3B-1%29%5Ccup+%28-1%3B0%29%5Ccup+%280%3B%2B%5Cinfty%29%0A%5C%5Cx_1%3D+%5Cfrac%7B2%2B%5Csqrt%7B4a%5E2%7D%7D%7B2%28a%2B1%29%7D+%3D+%5Cfrac%7B1%2B%7Ca%7C%7D%7Ba%2B1%7D+%0A%5C%5Cx_2%3D+%5Cfrac%7B1-%7Ca%7C%7D%7Ba%2B1%7D+)
3) при D=0 и a≠-1 имеет 2 совпадающих корня:
![4a^2=0 \\a=0 \\x_1=x_2= \frac{2}{2(a+1)} = \frac{1}{a+1}=\frac{1}{1}=1](https://tex.z-dn.net/?f=4a%5E2%3D0%0A%5C%5Ca%3D0%0A%5C%5Cx_1%3Dx_2%3D+%5Cfrac%7B2%7D%7B2%28a%2B1%29%7D+%3D+%5Cfrac%7B1%7D%7Ba%2B1%7D%3D%5Cfrac%7B1%7D%7B1%7D%3D1+)
4) при D<0 и a≠-1 не имеет корней
![4a^2\ \textless \ 0 \\a \in \varnothing](https://tex.z-dn.net/?f=4a%5E2%5C+%5Ctextless+%5C+0%0A%5C%5Ca+%5Cin+%5Cvarnothing)
дальше не рассматриваем этот случай
Ответ:
![a\in \{-1\} \cup \{0\} \Rightarrow x=1 \\a \in (-\infty;-1)\cup (-1;0)\cup (0;+\infty) \Rightarrow x_1=\frac{1+|a|}{a+1};\ x_2= \frac{1-|a|}{a+1}](https://tex.z-dn.net/?f=a%5Cin+%5C%7B-1%5C%7D+%5Ccup+%5C%7B0%5C%7D+%5CRightarrow+x%3D1+%5C%5Ca+%5Cin+%28-%5Cinfty%3B-1%29%5Ccup+%28-1%3B0%29%5Ccup+%280%3B%2B%5Cinfty%29+%5CRightarrow+x_1%3D%5Cfrac%7B1%2B%7Ca%7C%7D%7Ba%2B1%7D%3B%5C+x_2%3D+%5Cfrac%7B1-%7Ca%7C%7D%7Ba%2B1%7D)
2х²-3х-2≤0
Д=3²-4*2*(-2)=9+16=25=5²
х1=(3+5)/4=2
х2=(3-5)/4=-0,5
+. - +
_____.____._____
-0,5 2
[-0,5;2]
5(х-3) -2(х+7)=7
5х-15 - 2х -14=7
3х=7+29
3х=36
х=36:3
х=12
0,5-2х -(0,7х-2,1)=0,1-0,9(3х-1)
0,5 -2х - 0,7х+2,1=0,1 - 2,7х+0,9
-2,7х+2,6= -2,7х+1 - нет решений
10(2х -1)-3(4х-5)=66
20х - 10 - 12х +15=66
8х=61
х=61:8
х=7,625=7 5/8
5(5х-1)+0,2х=2,7х -6,5 -0,5х
25х -5+0,2х=2,2х -6,5
25,2х -2,2х=5-6,5
23х= - 1,5
х= -1,5:23
х= - 3/46