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Koshka6996
@Koshka6996
July 2022
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Знайти область визначення функції: у = 1 / √( - х2 + 3х + 18)
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Знайти область визначення функції: у = 1 / √( - х²
+ 3х + 18)
( - х2 + 3х + 18)>0
-(x²-2·3/2 x+9/4)+9/4+18= -(x-3/2)²+81/4
т.о. 0 < [- х2 + 3х + 18] ≤ 81/4
0 < √( - х² + 3х + 18) ≤ 9/2 ⇔ 1/(9/2) ≤ { 1 / √( - х² + 3х + 18)}<∞
2/9 ≤{ 1 / √( - х² + 3х + 18)}<∞
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Verified answer
Знайти область визначення функції: у = 1 / √( - х² + 3х + 18)( - х2 + 3х + 18)>0
-(x²-2·3/2 x+9/4)+9/4+18= -(x-3/2)²+81/4
т.о. 0 < [- х2 + 3х + 18] ≤ 81/4
0 < √( - х² + 3х + 18) ≤ 9/2 ⇔ 1/(9/2) ≤ { 1 / √( - х² + 3х + 18)}<∞
2/9 ≤{ 1 / √( - х² + 3х + 18)}<∞