[求助]模考数学求解

joanlq99

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present.  If the concentration of chemical B is increased by 100 percert, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
答案是40% decrease

If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?
答案是16

不解ing,求各位大虾帮忙看看...

summerice14

40% .....

Chemical = A/B^0.5

B2=2B and  Chemical is constant => A2=2^0.2 = 1.414A , thus A2 is around 40% more than A

龙少

rate与A平方成正比,与B成反比. RATE=kA平方/B  B增加100%,即成为2B,所以保证rate不变,A必须变成原来的根号2倍.即1.414A.  答案为40%increase.  LZ答案错啦~~

joanlq99

答案确实是40%increase,是偶看错了

可是第二题还是不明...