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maxim162011
@maxim162011
June 2022
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Найти сумму целых решений неравенства [tex] x^{2} -10[/tex]≤[tex] x+ \sqrt{10} [/tex]
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sangers1959
Verified answer
X²-10≤x+√10
(x-√10)(x+√10)-(x+√10)≤0
(x+√10)(x-√10-1)≤0
x+√10=0 x=-√10≈-3,16
x-√10-1=0 x=√10+1≈4,16
-∞_____+____-3,16____-_____4,16____+____+∞
x∈[-3,16;4,16]
-3, -2, -1, 0, 1, 2, 3, 4 ⇒
∑=4.
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Answers & Comments
Verified answer
X²-10≤x+√10(x-√10)(x+√10)-(x+√10)≤0
(x+√10)(x-√10-1)≤0
x+√10=0 x=-√10≈-3,16
x-√10-1=0 x=√10+1≈4,16
-∞_____+____-3,16____-_____4,16____+____+∞
x∈[-3,16;4,16]
-3, -2, -1, 0, 1, 2, 3, 4 ⇒
∑=4.