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WilliamProk
@WilliamProk
September 2021
1
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Решите неравенство методом интервалов:
(2/(x-1))-(1/(x+1))>3
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oganesbagoyan
Verified answer
(2/(x-1)) -(1/(x+1) ) >3 ;
0>3 +1/(x+1) -2/(x-1) <0 ;* * *0>3 +1/(x+1) -2/(x-1) ⇔3 +1/(x+1) -2/(x-1) < 0 * **
(
3(x²-1) +x -1 -2(x+1)
)/
(x-1)(x+1) <0 ;
(3x²- x -6)/(x+1)(x-1) < 0 ;
3
(
x -(1-√73)/6
) (
x -(1+√73)/6
) /
(x+1)(x-3) <0
;
методом интервалов :
+ - + - +
------------ (1 -√73)/6 ----------- (-1) ------------- (1) -------- (1+√73)/6 ----------
ответ : x∈
(
(1 -√73)/6 ; -1
)
U
(
1 ; (1 +√73)/6
) .
2 votes
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Answers & Comments
Verified answer
(2/(x-1)) -(1/(x+1) ) >3 ;0>3 +1/(x+1) -2/(x-1) <0 ;* * *0>3 +1/(x+1) -2/(x-1) ⇔3 +1/(x+1) -2/(x-1) < 0 * **
(3(x²-1) +x -1 -2(x+1) )/ (x-1)(x+1) <0 ;
(3x²- x -6)/(x+1)(x-1) < 0 ;
3(x -(1-√73)/6 ) ( x -(1+√73)/6 ) / (x+1)(x-3) <0 ;
методом интервалов :
+ - + - +
------------ (1 -√73)/6 ----------- (-1) ------------- (1) -------- (1+√73)/6 ----------
ответ : x∈( (1 -√73)/6 ; -1) U (1 ; (1 +√73)/6 ) .