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LyazzatMn
@LyazzatMn
July 2022
1
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Прошу помогите!!даю 80 баллов
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sedinalana
Verified answer
1
ОДЗ
x²-x-6≤0
x1+x2=1 U x1*x2=-6
x1=-2 U x2=3
x∈[-2;3]
√(6+x-x²)/(2x+5)-√(6+x-x²)/(x+4)≥0
√(6+x-x²)(x+4-2x-5)/(2x+5)(x+4)≥0
√(6+x-x²)(-x-1)/(2x+5)(x+4)≥0
√(6+x-x²)(x+1)/(2x+5)(x+4)≤0
√(6+x-x²)≥0⇒(x+1)/(2x+5)(x+4)≤0
x=-1 x=-2,5 x=-4
_ + _ +
-----------(-4)-----------(-2,5)--------[-1]-------------
x<-4 U -2,5<x≤-1+ОДЗ
Ответ x∈[-2;-1]
2
tga=3
cos²a=1:(1+tg²a)=1:(1+9)=1/10
cosa=1/√10
sina=√(1-cos²a)=√(1-1/10)=3/√10
sin2a=2sinacosa=2*3/√10*1/√10=0,6
cos2a=2cos²a-1=2*1/10-1=-0,8
(2sin2a-3cos2a)/(4sin2a+5cos2a)=(1,2+2,4):(2,4-4)=3,6:(-1,6)=-2,25
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Answers & Comments
Verified answer
1ОДЗ
x²-x-6≤0
x1+x2=1 U x1*x2=-6
x1=-2 U x2=3
x∈[-2;3]
√(6+x-x²)/(2x+5)-√(6+x-x²)/(x+4)≥0
√(6+x-x²)(x+4-2x-5)/(2x+5)(x+4)≥0
√(6+x-x²)(-x-1)/(2x+5)(x+4)≥0
√(6+x-x²)(x+1)/(2x+5)(x+4)≤0
√(6+x-x²)≥0⇒(x+1)/(2x+5)(x+4)≤0
x=-1 x=-2,5 x=-4
_ + _ +
-----------(-4)-----------(-2,5)--------[-1]-------------
x<-4 U -2,5<x≤-1+ОДЗ
Ответ x∈[-2;-1]
2
tga=3
cos²a=1:(1+tg²a)=1:(1+9)=1/10
cosa=1/√10
sina=√(1-cos²a)=√(1-1/10)=3/√10
sin2a=2sinacosa=2*3/√10*1/√10=0,6
cos2a=2cos²a-1=2*1/10-1=-0,8
(2sin2a-3cos2a)/(4sin2a+5cos2a)=(1,2+2,4):(2,4-4)=3,6:(-1,6)=-2,25