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несибе2208
@несибе2208
July 2022
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помогите срочно!!! неравенство и тригонометрия
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sedinalana
Verified answer
1
ОДЗ
x²+x-12≤0
x1+x2=-1 U x18x2=-12
x1=-4 u x2=3
x∈[-4;3]
√(12-x-x²)/(2x-7)-√(12-x-x²)/(x-5)≤0
√(12-x-x²)(x-5-2x+7)/[(2x-7)(x-5)]≤0
√(12-x-x²)(2-x)/[(2x-7)(x-5)]≤0
√(12-x-x²)≥0⇒(2-x)/[(2x-7)(x-5)]≤0
x=2 x=3,5 x=5
+ _ + _
-------------[2]------------(3,5)-------(5)--------------
2≤x<3,5 U x>5 +ОДЗ
x∈[2;3]
2
tga=3
сos²a=1:(1+tg²a)=1:10=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=2/10-1=-0,8
(3sin2a-4cos2a)/(5cos2a-sin2a)=(1,8+3,2)/(-4-0,6)=5/(-4,6)=-25/23
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Answers & Comments
Verified answer
1ОДЗ
x²+x-12≤0
x1+x2=-1 U x18x2=-12
x1=-4 u x2=3
x∈[-4;3]
√(12-x-x²)/(2x-7)-√(12-x-x²)/(x-5)≤0
√(12-x-x²)(x-5-2x+7)/[(2x-7)(x-5)]≤0
√(12-x-x²)(2-x)/[(2x-7)(x-5)]≤0
√(12-x-x²)≥0⇒(2-x)/[(2x-7)(x-5)]≤0
x=2 x=3,5 x=5
+ _ + _
-------------[2]------------(3,5)-------(5)--------------
2≤x<3,5 U x>5 +ОДЗ
x∈[2;3]
2
tga=3
сos²a=1:(1+tg²a)=1:10=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=2/10-1=-0,8
(3sin2a-4cos2a)/(5cos2a-sin2a)=(1,8+3,2)/(-4-0,6)=5/(-4,6)=-25/23