Solve a calculation problem, the question is as follows, vain hope that the answer is correct, the e

a...b...c...d

bc=√2/2ab

s=2×bc+ab=（1+√2）ab

I think I can see it.

Or let the length of the team be 1, the speed of the team walking is a, the speed of the herald is b, and the time to start and arrive at the end of the team is t

Quantity sought=bt 1=bt

So there is the equation at=1,1 (b-a) +1 (b+a)=t, and substituting a=1 t gives bt -2bt-1=

The solution yields bt=1+2

Solution Because the time is the same, s is the length of the team, so the relative distance of the pawn is 2s, the relative speed is the average velocity, 2*(a-b)(a+b) (a-b)+(a+b)=(a 2-b 2) a so the time is 2sa (a 2-b 2)=s b (the distance of the team is s) then there is 2ab=a 2-b 2And because the question asks about the length of the team exactly the distance traveled by the team, then you can use the speed ratio = distance ratio. Then the substitution option, such as b 2, so there is a=2 b=1, then 2*2*1=4-1 or so is not true, so only c is compliant.

The topic of civil servants is to choose not to do, and the options should be used.

Root number 2 Root number 3 (this question should be these options).

1. Combustion of charcoal, equation: c + o2 = = = = = combustion = = = = CO2 12 44

Let the burned charcoal be x grams: x grams generated by the reaction.

12 x = 44 x = grams.

The purity of the charcoal = due to the introduction of the generated gas into the lime water, the beaker increases the weight grams, indicating that the combustion reaction of the charcoal to generate CO2 is a gram reaction as follows: Ca(OH)2 + CO2**********==CaCO3 + H2O

The lime water participating in the reaction is y grams

y74/y=44/ y=

Participate in the reaction of calcium hydroxide in grams.

The key to answering this question is to understand that the mass of lime aqueous solution weight gain = the quality of carbon dioxide generated, and what other questions can be asked! Thanks for adopting.

Solution: (2x 3.)

3x 2 y-2xy 2

x 32xy 2

y 3 +（x 3

3x 2 y-y 3

2x 3 3x 2

y-2xy 2

x 3 2xy 2

y 3 x 3

3x 2 y-y 3

The result of Bu Qi after the simplification of 2y 3 has nothing to do with the imitation x, so Xiao Ming mistakenly copied x=2 into x=-2, but the result of his calculation is also correct.

Solution: Original = 2x 3

3x 2 y﹣2xy 2

x 3 2xy 2

y 3 Morino x 3

3x 2 y﹣y 3

2y 3 so the value of the original has nothing to do with the value of x, so when x=2011 is mistakenly copied as x= 2011, the value of the original is unchanged 2 (1) 3

My beginner Chinese level doesn't seem to be enough.

1. Solution y'=xy ==>dy y=xdx==>ln y =x 2 2+ln c (c is constant) ==>y=ce (x 2 2).

y=ce (x 2 2) is the solution of the original equation.

Obviously, y=0 is also the solution of the original equation, but it is contained in y=ce (x 2 2), so the general solution of the original equation is y=ce (x 2 2).

2. Solution: dy dx=-(2x+1)y 2

dy/y^2=-(2x+1)dx

Integral on both sides: -1 y=-x 2-x+c

1/y=x^2+x+c

y=1/(x^2+x+c)

3. Solution xlnydy=-ylnxdx

lny/y)dy=-(lnx/x)dx

2lny/y)dy=-(2lnx/x)dxln²y=-ln²x+c

ln²y+ln²x=c

I'll tell you when I've learned!