It is known that a and B are two unequal positive numbers, and a, x, y and B are in equal difference sequence in turn, and a, m, N and B are in equal proportion sequence in turn. Try to compare the size of X + y and M + n

It is known that a and B are two unequal positive numbers, and a, x, y and B are in equal difference sequence in turn, and a, m, N and B are in equal proportion sequence in turn. Try to compare the size of X + y and M + n

Let: A, x, y, B in turn into the observation of arithmetic sequence is D, then: x = a + D, y = a + 2D, B = a + 3D; a, m, N, B in turn into the common ratio of arithmetic sequence is Q, then: M = AQ, n = AQ ^ 2, B = AQ ^ 3, so there is a + 3D = AQ ^ 3, 3D = AQ ^ 3-A; because x + y = 2A + 3D = 2A + AQ ^ 3-A = a (1 + Q ^ 3) = a (1 + Q) (1 + Q + Q + Q ^ 2) m + n = AQ + AQ + AQ

It is known that a and B are two unequal positive numbers, and a, x, y and B are in equal difference sequence, a, m, N and B are in equal proportion sequence. Try to compare the size of X + y and
Compare the size of X + y and M + n

a. X, y, B in turn into arithmetic sequence
x+y=a+b
a. M, N and B are in equal proportion sequence
mn=ab
m+n>2√(mm)
m+n>2√(ab)
(x+y)-(m+n)
=a+b-(m+n)
0
(x+y)-(m+n)

If the five numbers 6, x, y, Z and 54 are in equal proportion sequence, then the value of real number x is

The two answers upstairs don't seem to be correct
y^2=6*54=324
Y = 18 note: y cannot be - 18
Q = ± root 3
When q = root 3
X = 6 root sign 3
When q = - radical 3
X = - 6 root sign 3
So x = 6 radical 3 or - 6 radical 3

Connect the photosensitive resistance R, setting resistance R0, ammeter, voltmeter, switch and power supply into the circuit as shown in the figure. The resistance of the photosensitive resistance decreases with the increase of the light intensity. Close the switch and gradually increase the light intensity of the photosensitive resistance. Observe the change of the indicator of the ammeter
A. Table a and V denote that the number becomes smaller, B. A denotes that the number becomes smaller, V denotes that the number becomes larger, C. A denotes that the number becomes larger, V denotes that the number becomes smaller, D. table a and V denote that the number becomes larger

Close the switch, photosensitive resistance R and constant resistance R0 in series, the total resistance in the circuit is equal to the sum of photosensitive resistance and constant resistance, the ammeter measures the current in the circuit, the voltmeter measures the voltage at both ends of the photosensitive resistance, gradually increase the light intensity of the photosensitive resistance, the resistance of the photosensitive resistance decreases, the total resistance of the circuit R total

(10-3/55×1)+(9-3/55×2)+(8-3/55×3)+…… +(2-3/55×9)+(1-3/55×10)=?
（10-3/55×1）+（9-3/55×2）+（8-3/55×3）+…… +（2-3/55×9）+（1-3/55×10）=?

Original formula = 10 + 9 + 8. + 1-3 / 55 × (1 + 2 + 3. + 10)
=55-3/55×55
=55-3
=52

The power supply voltage is 6V. When the switch S is closed, the indication of the ammeter is 0.5A. If the resistance value of resistance R2 is 6 Ω, 1) calculate the current and electric power through resistance R2
2) The total resistance of the circuit
3) The work done by the current to the resistor R1 within 10s
R1 and R2 are connected in parallel, and the ammeter measures the current of R1

1. The current through R2 is I2 = u / r2 = 6 / 6 = 1a
The electric power of R2 is P2 = ui2 = 6 * 1 = 6W
2.R1=U/I1=6/0.5=12Ω
1 / R union = 1 / R1 + 1 / r2 = 1 / 12 + 1 / 6 = 1 / 4
Then the total resistance in the circuit is R and = 4 Ω
3. The work done by the current to the resistance R1 within 10s of power on is as follows:
W1=UI1t=6*0.5*10=30J

The number of all integer solutions of equation (x2 + x-1) x + 3 = 1 is ()
A. 5 B. 4 C. 3 d. 2

(1) When x + 3 = 0, x 2 + X-1 ≠ 0, the solution is x = - 3; (2) when x 2 + X-1 = 1, the solution is x = - 2 or 1. (3) when x 2 + X-1 = - 1, x + 3 is even, the solution is x = - 1, so all integer solutions of the original equation are - 3, - 2, 1, - 1, a total of 4

The resistance of the test pen is 1100k ohm
With the neon tube is, how much is through the human body?

A: the resistance of the test pen is 1100k ohm, which is in series with the neon tube, because only in series can it protect the neon tube. If the resistance of the human body is 2000 Ω in dry environment, the current through the human body is I = u / r = 220 / (1100 + 2000) = 0.07a

[1-1 / 2 of x] divided by X & # 178; - 3-4 of X

[1/(x-2)-1]÷(3-x)/(x²-4)
=[(x+2)-(x²-4)]÷(3-x)
=(-x²+x+6)/(3-x)
=(x²-x-6)/(x-3)
=(x-3)(x+2)/(x-3)
=x+2
If you don't understand, I wish you a happy study!

The rated power of several electrical appliances in Figure 1 is the closest to 1000W ()
A. Electric fan B. calculator C. TV D. rice cooker

A. The power of electric fan is about 15W ~ 150W, which is not in line with the theme; B. the power of calculator is about 1.5 × 10-4w, which is not in line with the theme; C. the power of TV is about 130W, which is not in line with the theme; D. the power of electric cooker is about 1000W, which is in line with the theme; so D