Help solve the math problem! Hurry to solve math problems!

2003(x-y)+2004(y-z)+2005(z-x)=y-2x+z=0 x=(y+z)/2

and 2005 (z-x) = -2003 (x-y) -2004 (y-z) into 2003 2 (x-y) + 2004 2 (y-z) + 2005 2 (z-x) = 2004

Simplifying -2*2003(x-y)-2004(y-z)=2004 Substituting x=(y+z) 2 into the above equation gives z-y=20042) The square number and absolute value are both non-negative, and the sum is zero, so the square number and the absolute value are zero.

x-3y+z=0 5x-4y+z=0

y=4x, z=-11x

xyz is not equal to 0, so x, y, z are all non-zero.

xy+yz+zx)/(x^2+y^2+z^2)=[(4-44-11)x^2]/[(1+16+1210x^2]=-51/138=-17/46

If the speed of the passenger car is a kilometer per hour, the speed of the truck is B kilometers per hour, and the distance between A and B is m kilometers.

10a=15b，b=2a/3

m (a+b)]*a+96=80%*m, substituting b=2a3 into the solution, m=480

Author: 2009 Male.

1+cos2a=2(cosa) 2, a is the acute angle cosa = root number ((1+cos2a) 2) = 3 52. The perpendicular line of the BC extension cord is made through A, and the perpendicular foot is H

then cosa = dh ad=3 5

It can be seen that ah ad = root number (1-(3 5) 2) = 4 5 ah dh = 4 3

And because the angle c = 45 degrees.

ah/ch=1

ch/dh=4/3

bh+2)/(bh+1)=4/3

bh=2ah=bh+bc=2+2=4

1) (cosa)^2=(1+cos2a)/2=9/25。Because a is an acute angle, cosa > 0, so cosa = 3 5.

2) Construct the equation by using the high equality corresponding to angle c and angle a. Let the height of the BC edge be H, and the perpendicular line from A to the BC edge intersects the BC edge with E, and let ED=X. then tanc*(1+x)=tana*x.

Because cosa=3 5, sina = 4 5. So tana=4 3, tanc=1.

Substituting the above equation yields: x=3So h=3*tana=(1+3)*tanc=4.

To do BC's high AE, then the angle AEC = 90 degrees, because the angle BAC=90 degrees, so the angle C=45 degrees, because the angle BAC=angle C=45 degrees, so AE=CE=CD+DE=1+DE, and because COSA=DE=DE=DE1+DE=4 5, so the high AE=4 on the edge of BC

Let y1=kx, y2=b x

y=kx+b/x²

Substitute x=2 with x=3 and y=19.

Solve k, b, values.

Then substitute x=-1 into y=kx+b x.

Get answers. I beg your pardon for being too hasty and the answer was not calculated.

Let y=kx+m x2, put x=2,3; y=19 generations come in and get k=5; m=36。That is, the original equation is: y=5x+36 x2. So when x=-1, y=31. Note: x2 is the square of x.

(1) First, the right side of the equation is divided into a(x-3), b(x+1) (x+1)(x-3), and according to the identity property, a(x-3), b(x+1), x+5 can be obtained

According to this equation, the system of equations a-b=1 and -3a b=5 can be listed and solved to obtain a=-1, b=-2

2) The average unit price, the cost of two purchases and the quantity of two purchases. Let the first unit price be A, and the second unit price be B.

The average unit price of A is 1000a+1000b, 1000+1000, simplified to A+b 2

The average unit price of B is 800+800 (800 a+800 b), which is simplified to 2ab a+b

To judge whose purchase method is more cost-effective is to see whose average unit price is low, with the unit price of A and the unit price of B, if "0, then the unit price of B is low.

If < 0, the unit price of A is low. (a+b 2) (2ab a+b) is simplified after the division to obtain (a-b) 2 2(a+b) because according to the actual a>0, b>0, so the original formula is also "0, so the unit price of B is low, and B's purchase method is more cost-effective.

1.Let's say a total of x eggs are laid.

Then the first hatch is less than two-thirds of five, that is, 2x 3 -5 eggs, and the second time the remaining one-half is less than 3 eggs, that is, (x-(2x 3 -5))*1 2)-3=x 6-1 2 eggs.

Since the number of eggs is a natural number.

So the first hatch of 2x 3 -5>0==>x> and x is divisible by 3, and the second hatch x 6-1 2>0===>x<12 combined min.

1.If there are x tons of water in the large pool and y tons of water in the small pool, it can be known by the conditions, the water in the large pool is x+y-10 tons, and the water in the small pool is x+y-20 tons, and the volume of the large pool is times that of the small pool

x+y-10=(x+y-20), taking x+y as an unknown, the equation is the water shared by the two pools, and the answer is 70 tons

2.The first condition is that the other 3 people are counted as 13 copies, and A produces 2 copies, so A produces two-fifteenths of the total, and in the same way, B produces one-fifth of the total, and C produces four-fifteenths of the total4, so A, B, and C produce three-fifths of the total, so D produces two-fifths of the total, and knows that D produces 60 parts, so the four produce a total of 150 parts

Oh, that's right

1。If the capacity of the large pool is 6x and the capacity of the small pool is 5x, then.

5x+20=6x+10

x=10, so the beam pond has a total of 110 tons of water.

2。Let A, B, C and D produce 2x, 3x, 4x, 6x parts respectively and produce 60 parts, then x=10

So a total of 150 parts were produced.

The first question is: a total of 70 tons, 30 tons small and 40 tons large! The second question was not calculated!

oo black tea dumplings oo

Correct explanation, I will not repeat it here.

16 meters, 18 meters, 24 meters.

The maximum common divisor is 2 meters.

8 + 9 + 12 = 29 segments.

（1）a^2-1；a^3-1；a^4-1；a 100-1 (the first 3 can be calculated directly, and the last one is guessed).

To apply the formula of 1 guess, 2-1 = 1, so the result of the calculation is the same as the result of the original calculation).

Because (a 5+a 4+a 3+a 2+a+1)=0, (a-1)(a 5+a 4+a 3+a 2+a+1)=0

A 6-1 = 0 and a 6 = 1 are obtained

(1)a^2-1;a^3-1;a^4-1.(This is just taken apart), (2) 2 199+2 198+2 197+......2 2+2+1 will be it (2-1), the original formula does not change, as known by the above law, the original formula = a 100-1

Again, multiply by (a-1), (a-1) (a 5 + a 4 + a 3 + a 2 + a + 1) = 0

A 6-1 = 0

So a 6 = 1

Hope it helps, o(o