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Anonymous users2024-02-07Knowing a 0, b 0, a + b = 1, then (1 a -1) (1 b -1) is the maximum value ?
1/a²-1)(1/b²-1)
1-a²)(1-b²)/a²b²
1-a)(1-b)][1+a)(1+b)]/a²b²ba[1+(a+b)+ab]/a²b²
2+ab)/ab
2/(ab)+1
2/(a-a²)+1
Let y=a-a, y =1-2a, so that y = 0 gets the only standing point a=1 2, when a 1 2, y 0, when a 1 2, y 0, so a=1 2 is the maximum point, which is also the maximum point, the maximum value of y = 1 2-(1 2) =1 4, y has no minimum value (infinitesimal value), so the minimum value of [2 (a-a)+1] = 2 (1 4) +1 = 3, 2 (a-a )+1 has no maximum value (infinity), and the minimum value of (1 a -1) (1 b -1) is 3, There is no maximum.
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Anonymous users2024-02-06The answer is (1,+ the main knowledge points tested, the judgment conditions of the derivable must be continuous and derivable (the left and right derivatives exist and are equal), and the infinitesimal quantity will be multiplied by the bounded quantity in the calculation process to be infinitesimal quantity, as shown in the figure.
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Anonymous users2024-02-05Definition of station: For a function, it refers to a point where the first derivative of the function is 0 (stationary point is also called a stable point, a critical point). For multivariate functions, a stationary point is a point where all first-order partial derivatives are zero. Therefore, the stationing points of these two multivariate functions are shown in the figure below.
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Anonymous users2024-02-04As shown in the figure below, this problem involves anomalous integrals.
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Anonymous users2024-02-03Find the derivatives on both sides first, and then find f'(x), and then integrals to find f(x):
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Anonymous users2024-02-02m p n is the easiest way to do it.
Special. a=2
b=1 to compare the size.
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Anonymous users2024-02-01Let the ratio be q, then a2=a1*q, a3=a1*q 2, a6=a1*q 5
Then: 2a1+3a2=2a1+3a1*q=1,(a1*q 2) 2=9*(a1*q)*(a1*q 5).
Simplification: 2a1+3a1*q=1, q 4=9*q*q 5
Q 2 = 1 9 is obtained by Eq, because the terms of the proportional series an are positive, i.e., q>0, so Q = 1 3, and substituting into the formula gets: A1 = 1 3
So the general formula for an is an=a1*q (n-1)=(1 3)*(1 3) (n-1)=(1 3) n (n 1).
1/bn=-2/[n(n+1)]=-2[1/n-1/(n+1)]
sn=-2[1-1/2+1/2-1/3+..1/n-1/(n+1)]=-2[1-1/(n+1)]=-2n/(n+1)
This can be solved using inequalities.
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