[tex] \left \{ {{x+1 \geq \frac{1}{1-x} } \atop {|x-1|+|x+1| \leq x+2}} \right. [/tex]. Created by sergeysargsyanotl7bg. algebra-ru

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sergeysargsyanotl7bg @sergeysargsyanotl7bg

August 2022 2 18 Report

[tex] \left \{ {{x+1 \geq \frac{1}{1-x} } \atop {|x-1|+|x+1| \leq x+2}} \right. [/tex]

sedinalana

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Решим 1
(x+1)+1/(x-1)≥0
(x²-1+1)/(x-1)(x+1)≥0
x²/(x+1)(x-1)≥0
x=0 x=1 x=-1
+ _ _ +
---------------(-1)------------[0]---------------(1)---------------
x<-1U x>1 U x=0
x∈(-∞;-1) U (1;∞) U {0}
Решим 2
1)x<-1
-x+1-x-1≤x+2
x+2x≥-2
x≥-2/3
нет решения
2)-1≤x≤1
-x+1+x+1≤x+2
x≥0
0≤x≤1
3)x>1
x-1+x+1≤x+2
2x-x≤2
x≤2
1<x≤2
x∈[0;2]
\\\\\\\\\\\ /////////////////////////
-------(-1)---------[0]----------(1)-----------[2]-----
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Ответ x∈(1;2] U {0}

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