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July 2022
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Решите неравенство
а) 5(х+1)-х>2х+13
б) 5х^2-11х+6>0
в) х^2+10х/х-5<0
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Verified answer
Решите неравенствоа) 5(х+1)-х>2х+13
б) 5х^2-11х+6>0
в) х^2+10х/(х-5)<0
а) 5(х+1)-х>2х+13 ⇔ ) 5х-х-2х>13-5 ⇔2x>8⇔x>4
б) 5х^2-11х+6>0
5х^2-11х+6=0
x1=[11-√(121-120)]/10=(11-1)/10=1 x2=[11+√(121-120)]/10=(11+1)/10=6/5 + - +
///////////////////1--------------------6/5//////////////////////////
x∈(-∞,1)∪(6/5,+∞)
в) х^2+10х/(х-5)<0 ⇔
x[x(x-5)+10]/(x-5)<0 ⇔x(x²-5x+10)/(x-5)<0 ⇔
т.к.
(x²-5x+10)=0
D=25-4·10<0 НЕТ КОРНЕЙ,
⇒(x²-5x+10)> 0 при всех x,
x(x²-5x+10)/(x-5)<0 ⇔
+ - +
---------0////////////////////////5------------------
x∈(0;5)
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Verified answer
Решите неравенствоа) 5(х+1)-х>2х+13б) 5х^2-11х+6>0
в) х^2+10х/(х-5)<0
а) 5(х+1)-х>2х+13 ⇔ ) 5х-х-2х>13-5 ⇔2x>8⇔x>4
б) 5х^2-11х+6>0
5х^2-11х+6=0
x1=[11-√(121-120)]/10=(11-1)/10=1 x2=[11+√(121-120)]/10=(11+1)/10=6/5 + - +
///////////////////1--------------------6/5//////////////////////////
x∈(-∞,1)∪(6/5,+∞)
в) х^2+10х/(х-5)<0 ⇔
x[x(x-5)+10]/(x-5)<0 ⇔x(x²-5x+10)/(x-5)<0 ⇔
т.к.
(x²-5x+10)=0
D=25-4·10<0 НЕТ КОРНЕЙ,
⇒(x²-5x+10)> 0 при всех x,
x(x²-5x+10)/(x-5)<0 ⇔
+ - +
---------0////////////////////////5------------------
x∈(0;5)