Fractional math problems for junior 2 and 2 fractional math problems for junior 2

Let the distance be 1, the time taken by A is t, and the time taken by B is t'。

then t = (1 2) a + 1 2) b = (a+b) 2ab,t'/2）*a + t'2) *b = 1, i.e. t'=2/（a+b）

t-t'=(a+b)/2ab - 2/(a+b）= [(a+b)^2 - 4ab]/ 2ab(a+b) = (a-b)^2 / 2ab(a+b) >= 0

So t>=t'

So when a=b arrives at the same time, when a is not equal to b B B arrives first.

Solution: Let the distance between the two places of ab be s, and the time required for b is to walk half of the distance of t and a half of the distance at a speed of a kilometer per hour, and the other half of the distance to walk at a speed of b kilometers per hour"

So the time of A.

Half of the time the speed is a kilometer per hour, and the other half of the time is a kilometer per hour"

So t=2s (a+b).

The time difference between A and B is (

Simplify it. Because a>0,b>0,s>0,(a-b) 2>0, (a-b) 2*s 2ab(a+b) 0, the time taken by A is greater than that of B.

So B arrives first.

Let the total distance be 1, then time A is 1 2a+1 2b

Let time t be at 2+bt 2=1 then t=2 (a+b).

Compare 1 2a+1 2b-2 (a+b)=(a-b) 2ab(a+b) Because a is not equal to b, the result is greater than zero, so B arrives first.

It's B who comes first. Let the total distance be x, the time used by A is t(a), and the time used by B is t(B)t(A)=x (2a)+x (2b)=(a+b)x (2ab)a*t(B) 2+b*t(B) 2=x

So t(b)=2x (a+b).

t(a)-t(b)=(a+b)x(2ab)-2x(a+b) to determine the magnitude relationship between this equation and 0.

X, A+B, and 2AB are all greater than zero.

So the original formula can be changed to (a+b)(a+b)-4ab in relation to the size of zero.

a*a+2ab+b*b-4ab=a*a-2ab+b*b=(a-b)(a-b) This formula is equal to zero.

That is, when a=b, A and B arrive at the same time, otherwise B arrives first.

Solution: Suppose the distance is supposed. (s is greater than 0).

TA=s 2b+s 2a=s(a+b) 2abtB=2s (a+b).

tA-tB=s(a+b) 2ab-2s (a+b)s((a+b) 2-4ab) 2ab(a+b)s(a-b) 2 2ab(a+b).

1. If a=b, then tA = tB, arrive at the same time.

2. If A is not equal to B, then T A is greater than T B, so B arrives first.

Let the total distance be S, the time taken by A is one-half S divided by A plus one-half S divided by B, B travels at half of the speed of A+B in the whole process, A needs the total time (A+B) S divided by 2AB B uses the total time 2S divided by A+B and then discusses the situation.

Solution to the second problem: If the original plan takes x months, the efficiency is 1 x

According to the title: (1 x)*(1+

Question 1: A 2b 3· (ab 2) -2 = a 2b 3 (a 2b 4)=1 b

Question 2: x 2-16 x 2+6x+16 + x x-4=

Question 3: (pq 2r) 3 2p r 2 + 1 2q = (p 3q 3 8r 3) 2p r 2 + 1 2q (p 2q 3 16r) +1 2q

p 2q 4 16rq) +8r 16rq=(p 2q 4+8r) 16rq, question 4: 1 (2x + 1-x 2 x)=1 [(-x 2+2x 2 + 1) x]=x (x 2++1).

Question 5: a-b a (a - 2ab-b 2 a)=(a-b) a [(a 2 - 2ab+b 2) a]=(a-b) a [(a-b) 2=1 (a-b),

Question 1: A 2b 3· (ab 2) -2 =a b (1 a b 4)=1 b

Question 2: x 2-16 x 2+8x+16+x x-4

x-4)(x+4)/(x+4)²+x/(x-4)

(x-4)²+x(x+4)]/(x²-16)

x²-8x+16+x²+4x)/(x²-16)

2x²-4x+16)/(x²-16)

Question 3: (pq 2r) 3 2p r 2 +1 2q=p q (8r ) 2p r )=p q 4r

Question 4: 1 (2x + 1-x 2 x)=1 [-x-1) ] x=-x (x-1).

Question 5: a-b a (a -2ab+b 2 a) = (a-b) a (a-b) a=1 (a-b).