The math master comes in and helps with a math problem, helps with a math problem The math master co

1. q w directly and the known number form a number There is no solution to this situation.

2. q w is an independent number and the relationship between numbers is multiplication.

That is: q*3-3*q*2=-3 w*3-3*w*2=-1 q=1 w=1 3 q+w=4 3

3. The algorithm of power q*q*q-3*q*q=-3 w*w*w-3*w*w=-1

This answer is a bit more complicated, and it's hard to write.

But the assumptions upstairs are wonderful, and I'd like to say I'm going to solve them directly.

Let q=x+1; w=y+1

x+1)^3-3(x+1)^2+3=0; (y+1)^3-3(y+1)^2+1=0

Simplify and eliminate x 2 and y 2 items.

x^3-3x+1=0; y^3-3y-1=0

The above two formulas are subtracted and added respectively.

x^3-y^3)-3(x-y)=-2; (x^3+y^3)-3(x+y)=0

x-y)(x^2+y^2+xy-3)=-2; (x+y)(x^2+y^2+xy-3)=0

Therefore, it can be judged that (x 2 + y 2 + xy-3) is not zero, i.e. x + y = 0

q+w=2 adds: I haven't been exposed to mathematics for too long after graduation, and there is something wrong with factorization, which should indeed be +xy, and it is now corrected. (x 2+y 2+xy-3) is not zero because if it is zero, then (x-y)(x 2+y 2+xy-3)=-2 is not true, so it is judged to be non-zero.

If we let the equation be zero, e.g. x=y=1, and bring back the original q=w=2, the original equation does not hold.

At first glance, the second floor makes sense

But if you think about it, his key step "(x 2+y 2-xy-3) is not zero" is very problematic, in fact, look at this thing, as long as it can be transformed into (x-y) 2, it can definitely be equal to zero, how to turn it? As long as x*y=3 is fine

That is, it is very simple, as long as x=y=root number 3, the second floor is not true, so according to the inference of the second floor, q+w should be equal to "2" or equal to "2 root number 3 +2".

Excuse me, but on the 2nd floor this (x 3-y 3) -3 (x-y) = -2; (x-y)(x^2+y^2-xy-3)=-2;There is a question.

However, (x 3-y 3)-3(x-y)=-2 should be (x-y)(x 2+y 2+xy-3)=-2 parentheses should be +xy, not -xy. If so, you can't push it like that.

I don't really understand the title.

Add the exponents in each monomial in the polynomial to get the degree, and then choose the number with the highest number as the number of the polynomial

example:

x 3y + x 4y times: 5

x 2+z 2 times: 2

The numerical factor in a monomial is called the coefficient of this monomial. The exponent of all letters and the number of times called this monomial.

x y to the power of 5 m to the power of 3 m to the power of n to the power of 2 : just the monomial coefficient is constant; i.e.: 2

Index: 1+5+3+5=14

The 5th power unknown of x y is xy, so the number of unknowns is the sum of the number of xy, which is 5 5 = 10 times, and the coefficient is the constant term 5m 3n 5 2

πx(y^5)(m^3)(n^5)/2 。

1), the coefficients of the polynomials about x, y, m, n are 2 , and the order is 1+5+3+5=14.

2), the coefficients of the polynomial about x and y are (m 3) (n 5) 2 , and the order is 1+5=6 .

3), the coefficient of the polynomial with respect to m, n is x(y 5) 2 , and the order is 3+5=8 .

First of all, it is divided into 3 + 3 + 4 = 10 portions.

Each serving is 190, so the first and second are 3*190=570, and the third person is 4*190=760

1900 includes 3 + 3 + 4 = 10 copies.

Each serving is 1900 10 = 190

A: Each of them is .

Basket A sells more than basket B every day.

24 4 = 6 (kg).

The basket is sold every day.

6 (3-1) = 3 (kg).

The basket is sold every day.

3 3 = 9 (kilograms).

Suppose B has xkg, and sells ykg every day. Then A has x+24kg, and sells 3ykg per day.

x-4y=x+24-4(3y)

x-4y=x+24-12y

x offset -4y = 24-12y

8y=24y=3A: 3*3=9 (kg).

3,000 customers per day, A is (x+24) kg, B is x kg, A basket sells 3y kg per day, B per day, ykg, equation on both sides, eliminate x, solve y

If frame A has x kg, then box B has x-24 kg; If frame B sells for y kilograms per day, then frame A sells for 3y kilograms per day.

x-4*3y=x-24-4y

The solution is y=3, that is, the first frame sells 3*3=9 (kg) every day

Note: * is a multiplier sign.

Solution: First of all, according to the time before and after displayed on the clock, it can be seen that the clock has been 24 + 5 = 29 hours since the time of alignment.

The clock is 2 minutes slower per hour, so the total is 2*29=58 minutes slower, so the current standard time should be 5:58

After setting the clock to standard time at 12:00 noon on the 25th, I saw that the clock was 5:00 on the afternoon of the 26th.

That is, this clock has been gone for 29 hours. Because this clock is 2 minutes slower every hour. So 29 hours slow 2x29=58 minutes.

Standard time should be 5:58.

2 minutes slower per hour, is it difficult to slow 2 * 5 = 10 minutes for 5 hours This table is 17:00 The actual time is 10 minutes slower, then the actual time will be added 10 minutes 17:10 hope to adopt.

This clock has been gone for 29 hours. Because this clock is 2 minutes slower every hour. So 29 hours slow 2x29=58 minutes. Standard time should be 5:58.

At 12 o'clock on the 26th, 24 hours have passed, and then at 5 o'clock in the afternoon of the same day, a total of 29 hours have passed, and because the clock is not 2 minutes slower, then 29*2=58, the standard time is 5:58

Because the second question of the first type is miscalculated:

The first algorithm: 220 + 190 + 200 + 240 + 16 = 866 (886 you wrote).