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kedbi4
@kedbi4
August 2022
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Помогите с ходом решения
x^3+6x^2+((28x^2+2x-10)/(x-5))<=2
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sedinalana
Verified answer
x^3+6x^2+((28x^2+2x-10)/(x-5))<=2
(x^4+6x
³-5x³-30x²+28x²+2x-10)/(x-5)-2≤0
(x^4+x³-2x²+2x-10-2x+10)/(x-5)≤0
(x^4+x³-2x²)/(x-5)≤0
x²(x²+x-2)/(x-5)≤0
x²+x-2=0⇒x1+x2=-1 U x1*x2=-2⇒x1=-2 U x2=1
x²(x+2)(x-1)/(x-5)≤0
x=0 x=-2 x=1 x=5
_ + + _ +
-------------[-2]--------------[0]--------------[1]---------------(5)--------------
x∈(-∞;-2] U [1;5) U {0}
4 votes
Thanks 5
kedbi4
Ответ правильный! Но как же правило деления и умножения на переменную в неравенствах!?( Неужели я что не так понимаю(
sedinalana
по правилу решения методом интервалов
kedbi4
Огромное спасибо за помощь!)
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Answers & Comments
Verified answer
x^3+6x^2+((28x^2+2x-10)/(x-5))<=2(x^4+6x³-5x³-30x²+28x²+2x-10)/(x-5)-2≤0
(x^4+x³-2x²+2x-10-2x+10)/(x-5)≤0
(x^4+x³-2x²)/(x-5)≤0
x²(x²+x-2)/(x-5)≤0
x²+x-2=0⇒x1+x2=-1 U x1*x2=-2⇒x1=-2 U x2=1
x²(x+2)(x-1)/(x-5)≤0
x=0 x=-2 x=1 x=5
_ + + _ +
-------------[-2]--------------[0]--------------[1]---------------(5)--------------
x∈(-∞;-2] U [1;5) U {0}