ОДЗ:
2х²-13х+20>0 U x+7>0
D=9,√D=3,x=(13-3)/4=2,5 U x=(13+3)/4=4
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2,5 4
x<2,5 x>4 U x>-7⇒x∈(-7;2,5) U (4;≈)
1)log(2)(2x²-13x+20)-1≥0 U log(3)(x+7)<0<br>2x²-13x+20≥2 U x+7<1<br>2x²-13x+18≥0 U x<-6<br>D=25, √D=5 x=(13-5)/4=2 U x=(13+5)/4=4,5
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2 4,5
x≤2 x≥4,5 U x<-6⇒x∈(-≈;-6)<br>(-≈;-6) + (-7;2,5) U (4;≈)⇒x∈(-7;-6)
2)log(2)(2x²-13x+20)-1≤0 U log(3)(x+7)>0
2x²-13x+20≤2 U x+7>1
2x²-13x+18≤0 U x>-6
D=25, √D=5 x=(13-5)/4=2 U x=(13+5)/4=4,5
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2 4,5
2≤x≤4,5 U x>-6⇒x∈[2;4,5]
[2;4,5] + (-7;2,5) U (4;≈)⇒x∈[2;2,5) u (;4,5]
Ответ: х∈(-7;-6)U[2;2,5) u (;4,5]