A / (a ^ 2-9), A-1 / A ^ 2 + 6A + 9 2. X / 2 (x + 1), 1 / x ^ 2-x 3.1 / x ^ 2-4, X / 4-2x

A / (a ^ 2-9), A-1 / A ^ 2 + 6A + 9 2. X / 2 (x + 1), 1 / x ^ 2-x 3.1 / x ^ 2-4, X / 4-2x

General division 1, a / (a ^ 2-9) and (A-1) / A ^ 2 + 6A + 9, a / (a ^ 2-9) = A / (a + 3) (A-3), (A-1) (/ A ^ 2 + 6A + 9) = (A-1) / (a + 3) ^ 2 (a + 3) = (A-1) / (a + 3) (a + 3) (a + 3), common denominator (a + 3) (a + 3) (A-3), namely a / (a ^ 2-9) = A / (a + 3) (A-3) = a (a + 3) (a + 3) (A-3), (A-1) (/ A ^ 2 + 6A + 9)

1 / x square, 1 / 4-2x, 1 / x + 2

1 / x square = 2 (2 + x) (2-x) / 2x & # 178; (2 + x) (2-x)
1/4-2x=x²(2+x)/2x²(2+x)(2-x)
1/x+2=2x²(2-x)/2x²(2+x)(2-x)

1 / x ^ 2-4,1 / 4-2x, 1 / x + 2

1/（x^2-4）=2/[2（x-2）（x+2）] 1/（4-2x）=-（x+2）/[2（x-2）（x+2）] 1/（x+2）=2（x-2）/[2（x-2）（x+2）]

Xiaoxin is going to take part in the toy car race. He uses the circuit shown in the figure to select a motor with high energy conversion efficiency. Assuming that the voltage of the battery is constant, he first holds the rotating shaft of the motor with his hand so that it does not rotate. After closing the switch, he reads the reading of the ammeter as 2A. Then, when the motor rotates normally, he reads the reading of the ammeter as 0.6A When the toy motor rotates normally, the efficiency of converting electric energy into mechanical energy is______ ．

Resistance of motor: r = ui1, total power consumed by motor: P total = u × I2, heating power: P heat = I22 × ui1, power converted into kinetic energy: P dynamic = P total - P thermal = i1i2 − & nbsp; i22i1u, efficiency of motor during normal rotation: η = P dynamic, P total = I1 − i2i1 = 2A − 0.6a2a = 70%, so the answer is: 70%

Given x (x-1) - (X & # 178; - y) = - 3, find the value of 2xy-x & # 178; + 3-y & # 178

∵x(x-1)-(x²-y)=-3
∴x²-x-x²+y=-3
∴x-y=3
∴2xy-x²+3-y²
=-(x²-2x+y²)+3
=-(x-y)²+3
=-(3)²+3
=-9+3
=-6

If two 220 V 60 W electric lamps are connected in series on the 220 V line, the power consumption of each lamp is 0

The resistance of the bulb
R=U2/P=220V2/60W
Because the same bulb is connected in series in the circuit of 220 V, the partial voltage is 110 V and the bulb resistance remains unchanged
From P = U2 / r = 110v2 / 220V2 / 60W = 15W
It can also be inferred from P = U2 / r that the resistance of the bulb remains unchanged, the electric power of the bulb is proportional to the square of the voltage, the actual voltage becomes 1 / 2 of the rated voltage, and the actual power becomes 1 / 4 of the rated power

When x is the value, the value of the following fraction is the square of zero (1) x-3 / X-9

The square of x-3 / X-9 = 0
x/(x-3)=+-3
x1=9/2
x2=9/4
It's a fractional equation. Don't forget to test it

When a conductor is connected to a circuit, the resistance of the conductor and the voltage at both ends will be ()
A. The resistance and voltage become half of the original. B. the resistance remains unchanged and the voltage decreases to half of the original. C. the resistance remains unchanged and the voltage becomes twice of the original. D. the resistance and voltage remain unchanged

The conductor resistance is determined by the conductor material, length and cross-sectional area, the current is reduced to half of the original, the conductor material, length and cross-sectional area of the conductor remain unchanged, the conductor resistance remains unchanged; ∵ I = ur, R remains unchanged, I becomes half of the original; ∵ from u = IR, the voltage is reduced to half of the original; so B is selected

Solving problems by quadratic radical
1-√3
——The standard solution of
1+√3

Original formula = (1 - √ 3) (√ 3-1) / [(√ 3 + 1) (√ 3-1)]
=-(√3-1)²/[(√3)²-1²]
=-(3-2√3+1)/(3-1)
=(2√3-4)/2
=√3-2

My family has an incandescent lamp is 100 watts, connected to 220 V AC, how much is its current (a)? What is the formula?
If it uses a multimeter to measure its resistance, how many ohm resistance should it have?

Just use p = UI to calculate