Factorization. 4a2-4b2 + 4bc-c2 =

Factorization. 4a2-4b2 + 4bc-c2 =

(50 points) let x 1 = 2.5, X (n + 1) = 2 / (3-xn)
If you are satisfied with the answer, add the remaining points

X(n+1)=2/(3-Xn)
X(n+1)-1=2/(3-Xn)-1=（Xn-1）/（3-Xn）
X(n+1)-2=2/(3-Xn)-2=2（Xn-2）/（3-Xn）
Let an = (XN-1) / (xn-2)
A(n+1)=An/2,
Also: A1 = 3, so: an = 3 / 2 ^ (n-1)
An=(Xn-1)/(Xn-2)=3/2^(n-1)
Sorting: xn = (1-3 / 2 ^ (n-2)) / (1-3 / 2 ^ (n-1))

A rectangular kitchen to shop square tiles, need to choose how many decimeters of brick side length, just shop that is neat and economical?
The length of the rectangle is 30dm, the width is 24dm, and the side length of the floor tile requires integral decimeter

Landlord, I'll come
The greatest common factor of 30 and 24 is 6, and then the side length is 6. Believe me
I'm working on it, too
You first say: the greatest common factor of 30 and 24 is 6, and then use 30 × 24 = 720dm & # 178;, 6 × 6 = 36. This is the area of square brick 720 ／ 36 = 20, which can be divided exactly, which is neat and economical
Landlord

Questions about Kirchhoff's law of voltage
I have some doubts about KVL: 1. It is said in the book that there is no closed circuit in which two voltage sources with the same direction and different voltages are connected in series, because it does not meet KVL, but it is completely possible to make such a circuit in actual experiments, as long as the two voltage sources take a relatively small value and do not burn the wire and power supply. Having such a circuit means that it does not meet KVL, But KVL should be applicable to any circuit
2. In a circuit in which only the capacitor and the voltage source are connected in series, the voltage at both ends of the capacitor is always less than the voltage source at the beginning of charging. KVL is not applicable at this time, but KVL is the application of energy conservation in the circuit. Why is KVL not applicable at this time?

1. The problem of two voltage sources: the voltage source mentioned here is the ideal voltage source (internal resistance = 0, wire resistance = 0), of course, it does not meet KVL. But your actual experiment is the actual voltage source and the actual wire (both resistance are non-zero)
2. Series circuit of capacitor and voltage source: for the same reason, capacitor and power source are ideal components (internal resistance and wire resistance are all zero), not actual components. Capacitor voltage can not jump. At the initial time of charging, capacitor voltage is zero (set to zero initial), while power source is constant voltage source, so there must be a non-zero value. So KVL is not applicable
In order to understand this kind of problem, we should pay attention to the difference between ideal components and actual components. In the actual circuit (in the lumped parameter circuit), there will not be the problem that KVL is not applicable

The domain of the function y = arcsinx-2 / 2 is

The domain of definition of anti sine function is [- 1,1]

Use two pieces of 8 cm in length and 4 cm in width to make a square. What's the perimeter? Make it a rectangle

Is the question ambiguous?
It could be down there
The square is 8 + 8 + 4 + 4 + 4 + 4 = 32
The rectangle is 8 + 8 + 8 + 8 + 4 + 4 = 40

Given a = 0.00.025 (14 zeros), B = 0.00.04 (16 zeros). What is the result of finding a * B?

A * b = 0.000.1 (30 zeros)

Given vector M = (2 √ 3sinx, 2cosx), vector n = (cosx, cosx), Let f (x) = vector m · vector n
Given the vector M = (2 √ 3sinx, 2cosx), vector n = (cosx, cosx), Let f (x) = vector m · vector n. (1) find the minimum positive period and range of F (x). (2) in △ ABC, when the opposite sides of angles a, B, C are a, B, C respectively, if f (A / 2) = 3 and a ^ = BC, it is a judgment of the shape of △ ABC

1. f(x)=2√3sinxcosx+2cos^2x
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
Minimum positive period T = 2 π / 2 = π
Range y ∈ [- 1,3]
2. f(A/2)=2sin(A+π/6)+1=3
sin(A+π/6)=1 A=π/3
Cosine theorem
cosA=(b^2+c^2-a^2)/2bc=1/2 a^2=bc
b^2+c^2-bc=bc
b^2+c^2-2bc=0
(b-c)^2=0
b=c A=π/3
A triangle is an equilateral triangle

A rectangle and a square have the same perimeter. It is known that the length of a rectangle is 8 cm and the width is 6 cm. How many cm are the sides of a square?

(8 + 6) × 2 / 4 = 14 × 2 / 4 = 28 / 4 = 7 (CM) a: the side length of a square is 7 cm

Calculation formula of circumference of cylinder and cone
Speed!!!

Find perimeter of solid figure__ O"…）
Circumference of the base circle of a cone: 2 π R (R is the radius of the base circle)
The circumference of the cone expansion: 2 π R + 2L (L is the generatrix of the cone)