30 questions on elementary two point calculation are very urgent!! I m going to finish writing the a

x/2y)^2*(y/2x)-[x/y^2)/（2y^2/x)]x^2/4y^2*(y/2x)-(x/y^2)*(x/2y^2)x/8y-x^2/2y^4

xy^3/8y^4-4x^2/8y^4

xy^3-4x^2)/(8y^4);

x+1)/x]*[2x/(x+1)]^2-[1/（x-1)-1/（x+1)]

x+1)/x]*[4x^2/(x+1)^2]-4x/(x+1)-2/(x-1)(x+1)4x(x-1)/(x-1)(x+1)-2/(x-1)(x+1)(4x^2-4x-2)/(x-1)(x+1).Set the original stipulation of x months.

At the same time, it is assumed that the total engineering quantity is: 1. Of course, if you set it to y, you can make an appointment in the end.

So the speed of the work of A: 1 x

B's engineering speed: 1 (x+6).

The units are all "amount per month").

Then column the equation:

1 x+1 (x+6)) 4+(1 (x+6)) x-4)=1 so the solution is: x=12

A: The original provision was December.

The one you wrote is a one-dimensional equation, not a fractional problem.

Original = a+b

The details are as follows. Primitive = the power of a (a-b) - the power of b (a-b) = (the power of a - the power of b) (a-b).

a+b)(a-b)/(a-b)

a+b

(20)x+1/x-1/x=x+1-1/x=x/x=1

21)a/b+1+2a/b+1-3a/b+1=a+2a-3a/b+1=0/b+1=0

Add and subtract with the same denominator, the denominator remains unchanged, and the numerator is added or subtracted.

1. Solution:

Because m≠4 so: mx-n=4x

m-4)x=n

x=n/(m-4)

When m is not equal to 4, the solution of the equation mx-n=4x is x=n (m a*10b (10a+10b)=100ab 10(a+b)=10ab (a+b)

The answer to question 2 is: 10

Welcome to it!

Look at it. Let B's speed be 3x, then A's speed is 2x, and the total length of the road, multiplied by 3x = because A walked an hour earlier and walked, so the remaining 4x is walked by A and B in the same time according to 2 than Fu Cover 3, and A walked 6 kilometers less, then 12x-8x/5 of the paragraph book - 8x=6

x is equal to kilometers per hour of lack of commotion, and road length multiplied by 3 times equals kilometers. I won't ask any more questions.

Rinse it once, and the cherry blossoms remain.

f(a)=1/(1+a^2)

It is divided into two parts.

A1 and A2, A1 A2 A, then the first section of the Jane Cong is rinsed with A1, then the remainder.

f(a1)=1/(1+a1^2)

Rinse with A2 again.

f(a2)=f(a1)/(1+a2^2)=1/[(1+a1^2)(1+a2^2)]

1 [1+a1 2+a2 2+a1 2a2 2]1 [1+(a1+a2) 2-2a1a2+(a1a2) 2]1 [1+a 2+(a1a2) 2-2a1a2] because. a1a2)^2>=2a1a2

So. f(a2)=1/[1+a^2+(a1a2)^2-2a1a2]≤1/(1+a^2)=f(a)

It's good to use the second option.

The first step is to write first by 1 x-1 y=3

The second step is to write 1 x-1 y=y-x xy=3, and the third step is to write y-x=3xy

Step 4 Write Substituting into a known algebraic formula.

2x-14xy-2y)/(x-2xy-y)=[2(x-y)-14xy]/[x-y)-2xy]

Step 5 is equal to (-6xy-14xy) (3xy-2xy) Step 6 is equal to (-20xy) (5xy).

The last is equal to 4

Help me get the best one.

Because 1 x-1 y=3, (y-x) xy=3y-x=3xy

x-y=-3xy

So 2x-14xy-2y=2(-3xy)-14xy=-20xyx-2xy-y=-2xy-3xy=-5xy, so the original = 4