1. For rational numbers, a new operation is defined: X ⁃ y = ax + by + XY, where a and B are constants, and on the right side of the equation are the usual addition and multiplication operations It is known that: 2 * 1 = 7, ((#; 3) * 3 = 3, find the value of one third of B 2. Please write out a system of linear equations of two variables so that its solution is x = 1, y = 2. What is the system of equations?

1. For rational numbers, a new operation is defined: X ⁃ y = ax + by + XY, where a and B are constants, and on the right side of the equation are the usual addition and multiplication operations It is known that: 2 * 1 = 7, ((#; 3) * 3 = 3, find the value of one third of B 2. Please write out a system of linear equations of two variables so that its solution is x = 1, y = 2. What is the system of equations?

1. X * y = ax + by + XY, where a and B are constants
2※1=7,(−3)※3=3,
2a+b+2=7
-3a+3b-9=3
a=-25/3,b=-13/3
1/3※b
=1/3※(-13/3)
=-25/9+169/13-13/9
=131/13
The equations are as follows
x+y=3
2x-y=0

For rational numbers, a new operation is defined: X * y = ax + by + XY, where a and B are constants, and the right side of the equation is the usual addition and multiplication operation,
Given 2 * 1 = 7, (- 3) * 3 = 3, find the value of one third * 6.kkk
Before 1:00 this afternoon

∵ x * y = ax + by + XY, 2 * 1 = 7, (- 3) * 3 = 3, ∵ {2A + B + 2 = 7 {- 3A + 3b-9 = 3} {2A + B = 5} {A-B = - 4} ① + ② 3A = 1A = 1 / 3, substituting a = 1 / 3 into ② to get 1 / 3-B = - 4B = 13 / 3 ∵ x * y = 1 / 3x + 13 / 3Y + XY ∵ 1 / 3 * 6 = 1 / 3 × 1 / 3 + 13 / 3 × 6 + 1 / 3 × 6 = 28 and 1 / 9 or 253 / 9

For the rational number x y, a new operation is defined: X * y = ax + by + XY, where AB is a constant, and the right side is the usual addition and multiplication operation
Friends,
On the right side of the equation are the usual addition and multiplication operations. Given that 2 * 1 = 7, (- 3) * 2 = 1, find the value of 1 / 3 * 6,

According to the meaning of the title
2a+b+2=7
-3a+2b-3=1
∴a=6/7
b=23/7
∴x*y=6x/7+23y/7+xy
∴1/3*6=2/7+138/7+2
=22

Are there X3 + Y3 = Z3, X4 + Y4 = Z4, X5 + Y5 = Z5 and X6 + y6 = Z6 or even xn + yn = Zn?
4, 5, 6, n are all at the top right of X, y, Z
XYZ can be a non-zero integer, a zero, or a fraction. If there is one or two examples

If XYZ are all non-zero integers, it's Fermat's theorem. There is no solution
If 0 is allowed
Of course, there are solutions, such as x = 0, y = Z
When x = 0, for example, 0 ^ 5 + 3 ^ 5 = 3 ^ 5
Or 0 ^ 4 + 3 ^ 4 = (- 3) ^ 4
If XYZ is a non-zero fraction, there is no solution
If it's irrational, it's solvable
For example, (√ 5) ^ 4 + (√ 12) ^ 4 = (- √ 13) ^ 4

Polynomial x 10 times - x 9 times y + x 8 times y 2 times - x 7 times y 3 times + Write the eighth term and the last term according to the law, and point out that the polynomial is several times and several terms

X10 times - x9 times y + X8 times Y2 times - X7 times Y3 times + X6 times y times - X5 times Y5 times + X4 times y6 times - X3 times Y7 times + x2 times Y8 times - xy9 times + Y10 times
Item 8: - X3 times, Y7 times
It's better to move items 10 times
It's 10 times 11

It is known that real numbers x, y, Z satisfy XY divided by X + y = - 2, YZ divided by Y + Z = 4 / 3, ZX divided by Z + x = - 4 / 3,
Find the value of XYZ divided by XY + YZ + ZX

Divide XY by X + y = - 2 and take the reciprocal on both sides to get 1 / x + 1 / y = - 1 / 2
Similarly, 1 / y + 1 / z = 3 / 4, 1 / Z + 1 / x = - 3 / 4
Add the three formulas and divide by 2. The reciprocal of the formula you want is 1 / x + 1 / y + 1 / Z

ABC + A & sup2; b-ab & sup2; - A + B-C, factorization,
Note: the highest value in the formula is square
If it is a mistake, please correct it

abc+a²b-ab²-a+b-c
=c(ab-1)+a(ab-1)-b(ab-1)
=(c+a-b)(ab-1)

Factorization: A & sup2; b-ab & sup2; + A & sup2; c-AC & sup2; - 3ABC + B & sup2; C + BC & sup2;
kuai

=(a2b-ab2-abc)+(a2c-ac2-abc)+(b2c+bc2-abc)
=ab(a-b-c)+ac(a-c-b)+bc(b+c-a)
=ab(a-b-c)+ac(a-c-b)-bc(a-b-c)
=(a-b-c)(ab+ac-bc)
For a long time

Factorization AB (C ^ 2-D ^ 2) - CD (a ^ 2-B ^ 2)

ab(c^2-d^2)-cd(a^2-b^2)
=abc^2-abd^2-a^2cd+b^2cd
=(abc^2-a^2cd)-abd^2+b^2cd
=ac(bc-ad)+bd(bc-ad)
=(bc-ad)(ac+bd)

A mathematical problem in grade two of junior high school (on factorization)
It is known that x is an odd number greater than 3
1. Factorization (x ^ 3-3x ^ 2) - (x-3)
2. Try to explain that (x ^ 3-3x ^ 2) - (x-3) can be divided by 8

1. (x ^ 3-3x ^ 2) - (x-3) = x ^ 2 (x-3) - (x-3) = (x-3) (x ^ 2-1) = (x-3) (x + 1) (x-1) 2, ∵ x > 3 is odd ∵ X-1 is an even number greater than 2, x + 1 is an even number greater than 4, and x-3 is an even number greater than 0 ∵ all contain factor 2