1/x(x+2)+1/x(x+2)(x+4)+1/(x+4)(x+6)+…… 1（x+2004）(x+2006)

1/x(x+2)+1/x(x+2)(x+4)+1/(x+4)(x+6)+…… 1（x+2004）(x+2006)

The second one has an X, right?
The original formula = [1 / X-1 / (x + 2)] / 2 + [1 / (x + 2) - 1 / (x + 4)] / 2 +... + [1 / (x + 2004) - 1 / (x + 2006)] / 2
=[1/x-1/(x+2006)]/2
=2005/2x(x+2006)

Calculate 1 / 2 * 4 + 1 / 4 * 6 + 1 / 6 * 8 +. + 1 / 2002 * 2004 + 1 / 2004 * 2006

1/2*4+1/4*6+1/6*8+.+1/2002*2004+1/2004*2006
=（1/2-1/4+1/4-1/6+1/6-1/8+…… +1/2002-1/2004+1/2004-1/2006）÷2
=（1/2-1/2006）÷2
=501/2006

If x is greater than or equal to 2, then x ^ 2 + 4x + 4 > = 0
Is it a true proposition? Is it a true proposition?

x^2+4x+4=（x+2）^2>=0
If x is greater than or equal to 2, then x ^ 2 + 4x + 4 > = 0 is a true proposition
Its you no proposition is x ^ 2 + 4x + 4

How to simplify the process? Thank you

√8/27=√2/3*√4/9==2/3*√2/3=2/3*√2*3/3.3=2/9√6

How to determine the sign of ABC in quadratic function

If the sign of a is positive or negative, the opening of the curve is upward a > 0 and downward A0
The axis of symmetry is less than 0 - B / 2a0 if the intersection point is below the x-axis

Simple operation: 1.2345 + 2.3451 + 3.4512 + 4.5123 + 5.1234=

1.2345+2.3451+3.4512+4.5123+5.1234
=1+2+3+4+5+0.1+0.2+0.3+0.4+0.5++0.01+0.02+0.03+0.04+0.05+0.001+0.002+0.003+0.004+0.005+0.0001+0.0002+0.0003+0.0004+0.0005
=10+1+0.1+0.01+0.001
=11.111

Given the set M = {- 1,1}, n = {X / (1 / 2) < 2 (x + 1) power < 4, X ∈ Z}, then m ∩ n

(1 / 2) < 2 (x + 1) power < 4, that is: 2 ^ (- 1)

-36*a*a*x+24abc-4*b*b*x

Should it be like this?
-36*a*a*x+24abx-4*b*b*x
=-4x(9a^2-6ab+b^2)
=-4x(3a-b)^2

The polynomial x ^ 4 + 4Y ^ 4 is factorized;

x^4+4y^4
=x^4+4y^4+4x^2y^2-4x^2y^2
=(x²+2y²)²-4x²y²
=(x²+2y²-2xy)(x²+2y²+2xy)

2/3x=80-x-10-5,

2/3x=80-x-10-5
2/3x+x=65
5/3x=65
x=39