There are two math algebra problems that I can t do, and I hope a master can help. Thank you!!!

1. Because the proof conditions are long, it is advisable to simplify them first, so the analysis method is used.

Certificates: a(1 b+1 c) + b(1 a+1 c) + c(1 b+1 c) + 3=0

That is: a(1 b+1 c) + b(1 a+1 c) + c(1 a+1 b) = -3

The following is deformed: a(1 a+1 b+1 c) + b(1 a+1 b+1 c) + c(1 a+1 b+1 c)-3=-3 (add a a, b b, c c on the left, multiply in, and subtract back to 3).

a(1/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0

That is: (a+b+c)(bc+ac+ab) (abc)=0

Because a+b+c=0, the original proposition is true.

a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)

1 (b+1+bc)+1 (c+1+ac)+1 (a+1+ab) [abc=1 brought in].

ac+1) (1+ac+c)+1 (a+1+ab) [The two 1s in the first fraction of the above equation are both brought in with abc

1-1/(a+1+ab)+1/(a+1+ab)

1) Bring a=-b-c to the left of the equation.

a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/c)+3(-b-c)(1/b+1/c)+b(1/a+1/c)+c(1/b+1/a)+3

2-b/c-c/b+b/a+b/c+c/b+c/a+3(b+c)/a +1

a/(ab+a+1)

a/(1/c+a+1)

ac/(ac+c+1)

b/(bc+b+1)

b/(1/a+b+1)

ab/(ab+a+1)

ab/(1/c+a+1)

abc/(ac+c+1)

1/(ac+a+1)

So the original formula = ac (ac + c + 1) + 1 (ac + a + 1) + c ( a + c + 1).

1.Evidence: Because ABC≠0 shows that a, b, and c are not 0

a(1 b+1 c)+b(1 a+1 c)+c(1 b+1 a)+3

a/b+a/c+b/a+b/c+c/b+c/a+b/b+a/a+c/c

a+b+c)/a+(a+b+c)/b+(a+b+c)/c

Because a+b+c=0

So the result of the above equation is 0

That is, a(1 b+1 c) + b(1 a+1 c) + c(1 b + 1 c) + 3 = 0

Note: You may have copied the formula by mistake, it should not be c(1 b+1 c), but c(1 b+1 a).

In the first question, a(1 b+1 c)+b(1 a+1 c)+c(1 b+1 a)+3=0, it is known that abc ≠ 0, so you can multiply abc at the same time around the equation, and then sort it out to get ac(a+b+c)+ab(a+b+c)+bc(a+b+c)=0, and it is known that (a+b+c)=0, so the equation is true.

If someone is faster than me, I won't answer.

The fractional numerator and denominator are both divided by xy:

(2 y +3-2 x) (1 y-2 1 x) because the condition 1 x -1 y=3 is known

So it is equal to -3 -5

Equal to 3 5 or write.

The fractional numerator and denominator are both divided by xy:

Equals: -3 5

(1): The triangle ADC is similar to the triangle BEC, so be AD=BC AC, so BE=9

1 2 angle abc + angle c + 60 = 180Angle abc + angle c + 45 = 180 So: angle b = 30 , angle c = 105

2. Pour the 3-liter bucket into the 7-liter bucket, repeat again, at this time, there are 6 liters of water in the 7-liter bucket, then fill the 3-liter bucket and pour it into the 7-liter bucket, there are 2 liters in the 3-liter bucket, pour the water from the 7-liter bucket into the 3-liter bucket, and leave 5 liters of water.

Question 1:

A: There are 15 ways to stand.

Question 2: Inverted:

1. Fill 7 liters first, 2. Pour 3 liters. (7 liters of the remaining 4 liters) 3, pour out the 3 liters. (3 liters empty).

4. Fill up 3 liters with 7 liters. (7 liters of slow width of the remaining liters) 5, pour out 3 liters. (3 liters empty).

6. Pour the remaining 1 liter into 3 liters. (There is 1 liter in 3 liters, and 7 liters of air front which brigade) 7, fill 7 liters, (7 liters are full).

8. Use 7 liters full of silver stools, fill 2 liters missing from 3 liters, and leave 5 liters of 7 liters.

Complete the task. It takes a total of 8 steps.

FWT good luck.

7-3=4, 4-3=1, 7-(3-1)=5, the reed rubber is filled with 7, poured full 3, poured 3, and used 7 to fill 3, left 1, poured into 3, then the remaining space of 3 is 2, and then filled with 7, and then use 7 to pour the remaining mu in 3 to search for the remaining space, so there are still 5 in 7

Only the first question is given, because the second question is correct.

6 students stand in a row of small chorus, Huahua Ruler is the lead singer, can not stand at both ends, how many different tactics are there?

CA480 (scum).

Set a load of x tons and B load of y tons.

2x+3y=

5x+6y=35

4x+6y=31

x=4，y=

Third gross weight: 3x+5y=, freight:

Solution: Set up the original male x and female Y people.

x(1+10%)+y(1-10%)=804x+y)(1+

The solution is x=420 y=380

Then the current number of boys is 420 (1+10%)=462The current number of girls is 380 (1-10%)=342Answer: The current number of boys is 462, and the current number of girls is 342.

Solution: Let the number of boys be x and the number of girls be y.

x+y=804

10%x-10%y=804-804/

x=422y=382

A: There are 422 boys and 382 girls.

123...The total number of sequences n.

1 is followed by n-1 numbers, and the sequential number is n-1.

The ordinal number of 2 is n-2.

The total number of strings in order is: 1+2+.n-1) = n(n-1) 2, and the inverse ordinal number is 0

Then the numbers inside are exchanged, and for each decrease in the ordinal number, the inverse number increases by one, so that their total remains the same, n(n-1) 2

Now back to your question, the reverse order is k, so the order is n(n-1) 2-k.

Since all the numbers are reversed, the original ordinal number becomes all in reverse order, and the reverse order is all in order, so the inverse ordinal number is .

n(n-1) 2 - kpcs.

Multiple choice, 1 (d).

2(a)3(a)bc looks the same.

4（b）5（d）

Fill-in-the-blank questions and 16

4. I don't know yet.

5. It is not equal to -2 or 1