Factorization x ^ 2-y ^ 2 + Y-1 / 4

Factorization x ^ 2-y ^ 2 + Y-1 / 4

X^2-Y^2+Y-1/4 =X^2-(Y^2-Y+(1/2)^2 =X^2-(Y-1/2)^2 =(x+y-1/2)(x-y+1/2)

Factorization 4 (X-Y) ^ 2-6 (Y-X) + 2

4(x-y)^2-6(y-x)+2
=[2(x-y)-1][2(x-y)-2]
=(2x-2y-1)(2x-y-2)
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Factorization of (x-1) (X-2) (x-3) (x-4) - 24

(x-1)(x-2)(x-3)(x-4)-24 =[(x-1)(x-4)][(x-2)(x-3)]-24=[(x^2-5x)+4][(x^2-5x)+6]-24=(x^2-5x)^2+10(x^2-5x)+24-24=(x^2-5x)^2+10(x^2-5x)=(x^2-5x)(x^2-5x+10)=x(x-5)(x^2-5x+10)

What is (x + 1) (x-3) + 1 equal to (factorization)
What is (x + 1) (x-3) + 1 equal to
Factorization
This is from our homework
hungry
I made a mistake and changed it to: (x-1) (x-3) + 1

(x-1)(x-3)+1
=x²-3x-x+3+1
=x²-4x+4
=(x-2)²

X-x of 2-x and 178; - 4x + 4-x of 4-x and 178;

X-x of 2-x and 178; - 4x + 4-x of 4-x and 178;
=-x/(x-2) +(x²-4)/(x²-4x+4)
=-x/(x-2) +(x+2)(x-2)/(x-2)²
=-x/(x-2) +(x+2)/(x-2)
=2/(x-2)

The working principle of the following electrical appliances is related to eddy current phenomenon ()
A. Hair dryer B. electromagnetic range C. refrigerator D. TV set

A: The electric hair dryer is made of the principle that the electrified conductor is forced to rotate in the magnetic field, which is the same as the principle of the motor, so a is wrong; B: the electromagnetic stove uses the periodic change of the alternating current in the coil to generate the induced current in the nearby conductor to heat the food, even if the eddy current principle is used to heat the food

(a-2b + 3C) & # - 178; - (a + 2b-3c) & # - 178; simple calculation

Use square difference formula or complete square formula to add points

A 40 W incandescent lamp can work normally for hours with 1 kwh electric energy
If the incandescent lamp can generate 5J light energy per second in normal operation, the luminous efficiency of the incandescent lamp in normal operation is
Well, one more question
I hope you can also send the process
Thank you very much

First question: T = w / P = 1kwh / 40W = 25h
Second question: its power is 40W, that is, it consumes 40J of electric energy per second. If the light produced is 5J, the efficiency is 5 / 40 = 12.5%

0.625*10*8*11*2*12

0.625*10*8*11*2*12
=(0.625*8)*2*10*11*12
=(5 / 8) * 2 * 10 * 11 * 12
=5*2*10*11*12
=10*10*11*12
=100*132
=13200

There are two resistors R1 and R2, which are 10 ohm and 20 ohm respectively. They are connected in series to a 60V circuit
Ask for:
(1) The amount of current in the circuit
(2) What are the voltages at both ends of R1 and R2?
(3) What is the work done by the current passing through R1 and R2 after 5 minutes of power on?
(4) Find out the ratio of work done by R1 and R2

1) I = u / r = 60V / (10 Ω + 20 Ω) = 2A
2) U1 = R1 * I = 10 Ω * 2A = 20V
U2 = R2 * I = 20 Ω * 2A = 40V
3）W1=UIT=20V*2A*300S=12000J
W2=UIT=40V*2A*300S=24000J
4)W1/W2=12000J/24000J=1/2