Junior high school questions Who can help me solve them, please ask my friends to help me solve a ju

1/a-1/(a+1)+.1 (A+2) = 1 A-1 (A+2) has a dollar.

a/[(x+y)/2]=2a/(x+y)--1)a/(2x)+a/(2y)=a(x+y)/(2xy)--2)f=[2a/(x+y)]*2xy/[a(x+y)]=4xy/(x+y)^2

It is as cost-effective as x=y.

When x+y < 2 roots (xy), the first one is cost-effective.

When x+y > 2 roots (xy), the second type is cost-effective.

a(a+1) 1/1 + (a+1)(a+2) 1/2 +....(A+2009)(A+2010).

1／a)-(1／a＋1)+(1／a＋1）……1／a+2009）-(1／a+2010)

1 a)-(1 a+2010)=2010 a(a+2010)Question 2:

First of all, crack terms. 1/a(A+1) = 1/1 of A minus 1/1 (A+1).

Similarly, (a+1)(a+2) 1 = (a+1) 1 minus (a+2) 1 part.

A+2009)(A+2010) 1 = (A+2009) 1 minus 1 (A+2010).

In this way, it can be eliminated before and after.

Finally = 1/a minus (a+2010) 1/1 = a(a+2010) 2010

Conversion unit Cubic decimeter to cubic meter Convert this yourself Let me tell you the idea of answering the question Calculation is not enough. And when you do that, you can figure out the volume of water (by mass kilogram divided by density), and subtract the volume of water from the volume of the container and the volume of the stone is the volume of the stone, because the water rises because the stone goes in.

And then the second question 25* gives the mass of the stone, and divides the mass by the volume of the stone on top is the density, thank you.

(1) This point is the point of symmetry of point A (or B) with respect to the line L and the intersection point of the line B (or A) with the line L. (Proved by the principle of symmetry and the shortest line segment between two points).

2) This point is the intersection of the line between points A and B and the line L. (Proven by the fact that the difference between the two sides of the triangle is less than the third side).

2 2 3 2) (4 2 1 2) 0, the two points A and B are on the same side as L, and only the intersection point of the straight line AB and L is required, that is, the point sought by (2);

To solve (1) simply make a symmetry point a with respect to l', or make b with respect to the symmetry point b of the straight line l'to find ab'The intersection with L or BA'and L.

The first p has two points (the symmetry point of two points is found separately with l as the mirror, and the point where the line determined by the symmetry point and another point intersects with l is the p point) and the second one has a value (the straight line determined by the two points intersects p with l). It's not that you don't know how to read, just read one side!

It is a question of finding the maximum and minimum between three points, the sum of the two sides of the triangle is greater than the third side, and the difference between the two sides is less than the third side. These three points can form a triangle, according to this theory can be solved, the specific calculation yourself, I didn't bring scratch paper.

Add z at the same time, we get: x+y+2z=4x+2y=3x, according to the latter equation, then 2y=-x, x:y=(-2):

1, and then substitute x=-2y into x+y+2z=3x, so 2z=-5y, so x:y:z=(-4):

3x-z=x+y+z...1)

3x-z=4x+2y-z...2)

There is (1) y=2x-2z 3) substituted (2) 3x-z=4x+2(2x-2z)-z

3x-z=4x+4x-4z-z

4z=5x =>x:z=4:5

x = 4 out of 5 z....Substitution (3).

y = 8 out of 5 z-2z = > 5y = 8z - 10z = > 5y = -2z = >y: z = -2:5

So x:y:z=4:-2:5.

Let x+y+z=k(1), 4x+2y-z=k(2), 3x-z=k(3), (2) (3) x+2y=0

x=-2yx:y=-2:1;

Bring x:y=-2:1 into (1).

z-y=k(4).

Bring in (2) to get -z-6y=k(5).

4)-(5)=2z+5y=0,2z=-5y,z:y=-5:2 because x:y=-2:1,z:y=-5:2

Multiply x:y=-2:1 to obtain x:y=-4:2

After merging, x:y:z=(-4):2:(-5) is obtained

The equation is reduced to x+y+z=4x+2y-z, and then from 3x-z=x+y+z, z=-(5 2)y, so x:y:

z=-2y:y:-(5/2)y=-2:

4x+2y-z=3x-z, then 4x+2y=3x, then x=2y, bring x+y+z=4x+2y-z, and get 2z=7y

then unify into x and then compare the result.

Think of one of the letters (e.g. y) as a constant and solve a system of binary equations.

x+y+z=3x-z

4x+2y-z=3x-z

x=-2y z=-5/2y

x：y：z=（-2y）：y：((5/2y)=(2)：1：（-5/2）=（4）：2：（-5）

r13=r1+r3

10 ohms. r 132 = r13r2 (r13+r2).

50 15 = 10 withered 3 ohms.

r1324=r132+r4

10 3 + 10 = 40 3 ohms.

r total = r1324r5 (r1324+r5) 40 7 ohms.

2）i5=u/r5=10/10=1a

3）u2:u4=r2:r4=5:10=1:2u2+u4=10v

3u2=10v

u2=10/3v

(1) Proof: Link OC

then OC CD

ad cd again

oc〃ad∠boc=∠bad

i.e. boc = bac + cad

and BOC=2 BAC

2∠bac=∠bac+∠cad

bac=∠cad

AC divides the bad