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medoeddi
@medoeddi
August 2022
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Помогите решить неравенство из второй части ЕГЭ!!!
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ЕкарныйБабай
(2/(x-4) + (x-4)/2)²≤25/4
одз x≠4
(x-4)/2=t
(1/t+t)²≤25/4
1/ 1/t + t ≤ 5/2
(2+2t²-5t)/2t≤0
D=25-16=9
t12=(5+-3)/4 = 2 1/2
(t-2)(2t-1)/2t≤0
--------- 0 +++++ (1/2) -------- (1)++++++++
t = (-00 0) [1/2 1]
(x-4)/2<0 x<4
(x-4)/2≥1/2 x-4≥1 x≥5
(x-4)/2≤1 x-4≤2 x≤6
x=(-∞ 4) U [5 6]
2/ 1/t + t ≥ -5/2
(2+2t²+5t)/2t≥0
D=25-16=9
t12=(-5+-3)/4 = - 2 -1/2
(t+2)(2t+1)/2t≥0
--------- (-2) +++++ (-1/2) -------- 0++++++++
t=[-2 -1/2] U (0 +∞)
(x-4)/2≥-2 x-4≥-4 x≥0
(x-4)/2≤-1/2 x-4≤-2 x≤2
(x-4)/2>0 x-4>0 x>4
x= [ 0 2] U (4 +∞)
пересекаес с
x=(-∞ 4) U [5 6]
ответ [0 2] U [5 6]
2 votes
Thanks 1
ЕкарныйБабай
маловато за с
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Answers & Comments
одз x≠4
(x-4)/2=t
(1/t+t)²≤25/4
1/ 1/t + t ≤ 5/2
(2+2t²-5t)/2t≤0
D=25-16=9
t12=(5+-3)/4 = 2 1/2
(t-2)(2t-1)/2t≤0
--------- 0 +++++ (1/2) -------- (1)++++++++
t = (-00 0) [1/2 1]
(x-4)/2<0 x<4
(x-4)/2≥1/2 x-4≥1 x≥5
(x-4)/2≤1 x-4≤2 x≤6
x=(-∞ 4) U [5 6]
2/ 1/t + t ≥ -5/2
(2+2t²+5t)/2t≥0
D=25-16=9
t12=(-5+-3)/4 = - 2 -1/2
(t+2)(2t+1)/2t≥0
--------- (-2) +++++ (-1/2) -------- 0++++++++
t=[-2 -1/2] U (0 +∞)
(x-4)/2≥-2 x-4≥-4 x≥0
(x-4)/2≤-1/2 x-4≤-2 x≤2
(x-4)/2>0 x-4>0 x>4
x= [ 0 2] U (4 +∞)
пересекаес с
x=(-∞ 4) U [5 6]
ответ [0 2] U [5 6]