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Dasha010100
@Dasha010100
October 2021
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Сумма целых решений неравенства (х-4)/х^2-5х+4≥2х-5/х-1
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sedinalana
Verified answer
X²-5x+4=0
x1+x2=5 U x1*x2=4⇒x1=4 U x2=1
(x-4)/(x-4)(x-1)-(2x-5)/(x-1)≥0
1/(x-1)-(2x-5)/(x-1)≥0,x≠4
(1-2x+5)/(x-1)≥0
(6-2x)/(x-1)≥0
x=3 x=1
1<x≤3
x∈(1;3]
x={2;3}-2 целых решения
1 votes
Thanks 1
Dasha010100
Почему входит 1, если изначально единица в заминателе по одз исключается?
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Answers & Comments
Verified answer
X²-5x+4=0x1+x2=5 U x1*x2=4⇒x1=4 U x2=1
(x-4)/(x-4)(x-1)-(2x-5)/(x-1)≥0
1/(x-1)-(2x-5)/(x-1)≥0,x≠4
(1-2x+5)/(x-1)≥0
(6-2x)/(x-1)≥0
x=3 x=1
1<x≤3
x∈(1;3]
x={2;3}-2 целых решения