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poloroslik
@poloroslik
November 2021
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решите пожалуйста 2 номер методом интервалов
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sedinalana
Verified answer
1
(2x+1)/(x-3)-1≤0
(2x+1-x+3)/(x-3)≤0
(x+4)/(x-3)≤0
x=-4 x=3
x∈[-4;3)
2
x²-5x+4=0
x1+x2=5 U x1*x2=4⇒x1=1 U x2=4
x/(x-4)+5/(x-1)+24/(x-1)(x-4)≤0
(x²-x+5x-20+24)/(x-1)(x-4)≤0
(x²+4x+4)/(x-1)(x-4)≤0
(x+2)²/(x-1)(x-4)≤0
x=-2 x=1 x=4
+ + _ +
-----------[-2]------------(1)-----------(4)-----------------
x∈(1;4) U {-2}
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poloroslik
Спасибо большое.
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Answers & Comments
Verified answer
1(2x+1)/(x-3)-1≤0
(2x+1-x+3)/(x-3)≤0
(x+4)/(x-3)≤0
x=-4 x=3
x∈[-4;3)
2
x²-5x+4=0
x1+x2=5 U x1*x2=4⇒x1=1 U x2=4
x/(x-4)+5/(x-1)+24/(x-1)(x-4)≤0
(x²-x+5x-20+24)/(x-1)(x-4)≤0
(x²+4x+4)/(x-1)(x-4)≤0
(x+2)²/(x-1)(x-4)≤0
x=-2 x=1 x=4
+ + _ +
-----------[-2]------------(1)-----------(4)-----------------
x∈(1;4) U {-2}