Let f (x) be defined as R, and x = K / 2, f (x + 1) = - 1 / F (x). If f (x) is an odd function, f (x) = 3 ^ x when 0 is less than X and less than 2 (1) Find f (2001 / 4) (2) When 2K + 1 / 2 is less than X and less than 2K + 1, find the analytic expression of F (x) (k belongs to Z) (3) Is there a positive integer k such that when 2K + 1 / 2 is less than X and less than 2K + 1, log3f (x) is greater than x ^ 2-kx-2k has a solution

Let f (x) be defined as R, and x = K / 2, f (x + 1) = - 1 / F (x). If f (x) is an odd function, f (x) = 3 ^ x when 0 is less than X and less than 2 (1) Find f (2001 / 4) (2) When 2K + 1 / 2 is less than X and less than 2K + 1, find the analytic expression of F (x) (k belongs to Z) (3) Is there a positive integer k such that when 2K + 1 / 2 is less than X and less than 2K + 1, log3f (x) is greater than x ^ 2-kx-2k has a solution

If f (x + 1) = - 1 / F (x) is calculated as a function of period 2, then f (2001 / 4) = f (1 / 4) brings in "when 0 is less than X and less than 2, f (x) = 3 ^ X" to get 3 ^ 1 / 4
The back of their own calculation. With a cycle is good

All items in the sequence {an} are arranged in the following table according to the rule that each row has more items than the previous row
a1
a2a3
a4a5a6
a7a8a9a10
Note that the number sequence {BN}, B1 = A1 = 1, which is composed of the first column number A1 A2 A4 A7,..., is the sum of the first n terms of the number sequence {BN}, and satisfies 2bn / (bnsn-sn2) = 1 (n is greater than or equal to 2)
(1) It is proved that the sequence {1 / Sn} is an arithmetic sequence, and the general term formula of the sequence {BN} is obtained;
(2) In the above table, if from the third row, the numbers in each row form an equal ratio sequence from left to right, and the common ratio is the same positive number, when A81=
-(4 / 91), find the sum of all items in line k (k is greater than or equal to 3) in the above table

Use brackets to represent subscripts. If you can see clearly, just talk about ideas, then you won't write the calculation process
(1) B [n] = s [n] - s [n-1], which is brought into 2B [n] / (B [n] * s [n] - s [n] ^ 2) = 1, the relationship between 1 / s [n] and 1 / s [n-1] can be obtained by transforming the equation. It is proved that {1 / s [n]} is an arithmetic sequence, and then the general term formula of the arithmetic sequence 1 / s [n] = f (n), s [n] = 1 / F (n), and the general term formula of B [n] = s [n] - s [n-1] = 1 / F (n) - 1 / F (n-1)
(2) Because the common ratio of each row is the same, and the value of the first element of each row has been found in the previous question, so the sum of all items in the k-th row needs to find out how many items there are in each row and the common ratio Q. it is not difficult to see that there are k items in the k-th row. The method of finding the common ratio is a [81] = - 4 / 91. Calculate the first element of a [81] in the row from the first question, and then use the formula of equal ratio sequence to find the common ratio

When the voltage at both ends of a fixed value resistor changes from 4V to 6V, the current increases by 0.2A, and the change of power of the resistor is? W

From 6 / R-4 / r = 0.2, the power P of R = 10 is 6 * 0.6-4 * 0.4 = 2

1-2sin²75°

1-2sin^275°
=1-2sin^2(180°+95°)
=1-2sin^2(95°)
=1-2cos^2(5°)
=sin^2(5°)-cos^2(5°)
=-cos10°

A student used the power supply, voltmeter, ammeter, sliding rheostat, key and some wires to measure the value of resistance Rx. All components were in good condition, the power supply voltage was 6.0V and unchanged, and the sliding rheostat was marked with "20 Ω & nbsp; When the key is closed, the student finds that the current indication varies from 0.20 a to 0.58 a in the process of moving the slide of the sliding rheostat. When the slide is at a position near the midpoint, the indication of the voltmeter and the ammeter are shown in FIG. 13 (a) and FIG. 13 (b) respectively. (1) please fill in the relevant columns of the experimental data table in the following table. (2) the number of the voltmeter and the ammeter According to the experimental equipment and phenomena observed by the student, fill in the relevant data in the table, and calculate the resistance value of Rx. (when calculating the resistance, the accuracy is 0.1 Ohm)

(1) According to the principle of measuring resistance by voltammetry, the voltmeter measures the voltage at both ends of the resistance to be measured, the ammeter measures the current passing through, and the resistance is calculated according to r = UI. During the experiment, it is necessary to measure the average value for many times to reduce the error. As shown in the table, the number of physical quantity experiment is voltage (V), current (a), resistance (Ω) Rx, and resistance (Ω) 1, 2, 3 (2) When the resistance of the sliding rheostat connected to the circuit is zero, the current I3 = 0.58a, then R3 = u3i3 = 6.0v0.58a ≈ 10.3 Ω; when the sliding rheostat slide to the maximum resistance position, the resistance Rx is in series with the maximum resistance of the sliding rheostat, the current I1 = 0.20a, the voltage at both ends of the sliding rheostat: u slide = i1r slide = 0.20a × 20 Ω = 4V, the voltage at both ends of the resistance Rx: U1 = U-U slide = 6v-4v = 2V, then R1 = u1i1 = 2v0.20a = 10 Ω; ∵ the range of current representation is 0.20 ~ 0.58 a, ∵ the range of ammeter in figure (b) is 0 ~ 0.6A, the division value is 0.02A, then I2 = 0.28a, ∵ the voltage of power supply is 6V, ∵ the range of voltmeter in figure (a) is 0 ~ 3V, the division value is 0.1V, then U2 = 2.7V, R2 = u2i2 = 2.7v0.28a ≈ 9.6 Ω; ∵ the resistance of Rx: RX = R1 + R2 + R33 = 10.0 Ω + 9.6 Ω + 10.3 Ω 3 ≈ The following table shows: physical quantity experiment serial number voltage (V) current (a) resistance (Ω) RX resistance (Ω) 12.00.20 10.0 10.0 2 2.7 0.28 9.6 3 6.0 0 0.58 10.3

-How to solve the equation x2 + 32x-100 = 56?
Wrong. The equation is - x2 + 32x-100 = 100

-x^2+32x-100=100
-x^2+32x-200=0
x^2-32x+200=0
(x-16)^2+200-256=0
(x-16)^2=56
X1 = 16 + radical 56
X2 = 16 - radical 56

As shown in the figure, the voltage at both ends of the power supply does not change, and the resistance value of resistance R1 is 2 Ω. Close the switch S. when the slide P of sliding rheostat is at point a, the indication value of voltmeter V1 is 4V, and that of voltmeter V2 is 10V. When the slide P of sliding rheostat is at point B, the indication value of voltmeter V1 is 8V, and that of voltmeter V2 is 11V. Then the resistance value of resistance R2 is 4V___ Ω．

According to the constant total voltage, the sum of the indication of voltmeter V2 and the voltage at both ends of R1 should be the power supply voltage. The sum of the indication of voltmeter V1 and the voltage at both ends of resistance R2 is also the power supply voltage. According to the equal power supply voltage, the following equation can be obtained: 10V + i1r1 = 11V + i2r1 ①4V+I1R2=8V+I2R2… ② It can be reduced to R1 (i1-i2) = 1V ③R2（I1-I2）=4V… ④ Substituting R1 = 2 Ω into ③ and ④, the solution is R2 = 8 Ω, so the answer is 8 Ω

Help a master to solve the problem of higher numbers f '' (x) is continuous on [a, b]
F '(x) is continuous on [a, b]. It is proved that there is an m such that the definite integral of F (x) on a (lower bound) is equal to 1 / 2 (B-A) f (1 / 2 A + 1 / 2 b) + 1 / 24 (B-A) ^ 3 F' (m)
Thank you very much.

Are you sure there is half of the first item in the title
Suppose f (x) = 3, is a constant function, then f "(m) = 0
No matter what the value of M is, there will be
Left = ∫ f (x) DX = 3 (B-A)
Right side = 3 / 2 * (B-A)
Obviously not, the contradiction lies in the coefficient 1 / 2 of the first term
I can be 99% sure that there is no half of the title. Besides the above example, I can also give the proof of the revised title
prove:
Taylor expansion of F (x) in (a + b) / 2 to two intermediate remainder terms
f(x)=f((a+b)/2)+f'((a+b)/2)*(x-(a+b)/2)+[f"(ε)/2]*(x-(a+b)/2)^2
The integral of F (x) on [a, b]
∫f(x)dx =∫f((a+b)/2)+f'((a+b)/2)*(x-(a+b)/2)+[f"(ε)/2]*(x-(a+b)/2)^2 dx
=(B-A) * f ((a + b) / 2) + ∫ [f "(ε) / 2] * (x - (a + b) / 2) ^ 2 DX (the second integral is 0)
Because F "(x) is continuous on [a, b], so P

A 220 V, 60 W incandescent lamp and a 220 V, 100 W incandescent lamp are connected in series, and then connected to the 380 V power supply. Try to calculate the total current
And the power absorbed by two incandescent lamps and the voltage on two incandescent lamps

60W: 60 / 220 = current 0.2727a resistance = 220 / 0.2727 = 807 Ω 100W: 100 / 220 = 0.4545a220 / 0.4545 = 484 Ω series resistance = 807 + 484 = 1291 total current = 380 / 1291 = 0.2943 total power = 111.85w distribution 60W: 111 * 807 / 1291 = 70W voltage = 380 * 807 / 1291 = 237.5v100w: = 42W electricity

899×99
7.4*9.9+0.74
9.75*89.6/9.75*9.75/89.5/9.75
99999*22222+33333*33334
30.7-0.12*2-0.76
99*99+9.9/0.1
3.15/4-2.15*0.25
1111*37+9999*7

（900-1）*（100-1）=90000-100-900+1=890017.4*（10-0.1）+0.74=74-0.74+0.74=74（9.75/9.75）*（9.75/9.75）*（89.6/89.6）=133333*（3*22222+33334）=33333*（66666+33334）=33333*100000=333330000030.7-（0.24+...