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dgkjh
@dgkjh
December 2021
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неравенство с модулями.есть ли способ кроме раскрытия скобок по интервалам?
( |x+4| - |x+2| )(x²+8x-7)≤0
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sedinalana
Verified answer
( |x+4| - |x+2| )(x²+8x-7)≤0
1)x<-4
(-x-4+x+2)(x²+8x-7)≤0
-2(x²+8x-7)≤0
x²+8x-7≥0
D=64+28=92
x1=(-8-2√23)/2=-4-√23
x2=-4+√23
x≤-4-√23 U x≥-4+√23
x∈(-∞;-4-√23]
2)-4≤x≤-2
(x+4+x+2)(x²+8x-7)≤0
(2x+6)(x²+8x-7)≤0
x=-3 x=-4-√23 x=-4+√23
_ + _ +
-------------[-4-√23]---[-4]----[-3]-----[-2]--------[-4+√23]---------------
////////////////////////
x∈[-3;-2]
3)x>-2
(x+4-x-2)(x²+8x-7)≤0
2(x²+8x-7)≤0
-4-√23≤x≤-4+√23
x∈(-2;-4+√23]
Ответ x∈(-∞;-4-√23] U [-3;-4+√23]
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Answers & Comments
Verified answer
( |x+4| - |x+2| )(x²+8x-7)≤01)x<-4
(-x-4+x+2)(x²+8x-7)≤0
-2(x²+8x-7)≤0
x²+8x-7≥0
D=64+28=92
x1=(-8-2√23)/2=-4-√23
x2=-4+√23
x≤-4-√23 U x≥-4+√23
x∈(-∞;-4-√23]
2)-4≤x≤-2
(x+4+x+2)(x²+8x-7)≤0
(2x+6)(x²+8x-7)≤0
x=-3 x=-4-√23 x=-4+√23
_ + _ +
-------------[-4-√23]---[-4]----[-3]-----[-2]--------[-4+√23]---------------
////////////////////////
x∈[-3;-2]
3)x>-2
(x+4-x-2)(x²+8x-7)≤0
2(x²+8x-7)≤0
-4-√23≤x≤-4+√23
x∈(-2;-4+√23]
Ответ x∈(-∞;-4-√23] U [-3;-4+√23]