Fellow math geniuses, help kneel and beg for answers

1) a2+a=0, a=0 or -1, (2a) 2+2a=0 or 2

2) 3 pcs., (7, 13, 14).

3) To have the largest number of squares that are not equal to each other, each square number must be as small as possible, 1 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2 + 7 2 + 8 2 + 9 2 = 285 (*.)

So 359 can only be expressed as the sum of squares of 9 positive integers that are not equal to each other.

The question is whether 9 can be reached, and to be able to achieve it, you need to adjust the 9 squares in (*).

First, consider replacing the square number n2 less than 10 with a square number m2 of not less than 10, and increase the value by 359-285 = 74, that is, m2 - n2 = 74 is required to have a positive integer solution, and.

Since 74 is easy to know, m + n is the same parity as m - n, and both should be even, so it is divisible by 4, but 74 is not divisible by 4, contradictory!Therefore, there is no integer m, n makes m2 - n2 =74 true. Therefore there is no square number m2 not less than 10 instead of a square number n2 less than 10, so that the value is increased by exactly 359-285 = 74

Since (102 + 112) - (82 + 92) = 100 + 121 64 81 = 7674So for two integers not less than 10 m1, m2, two integers not greater than 9 n1, n2, more.

For the sum of squares of more than 3 integers not less than 10, instead of the sum of squares of integers of the same number not greater than 9, the difference will increase by more than 74

In summary, 359 cannot be written as the sum of the squares of 9 unequal positive integers, so 359 can only be expressed as the sum of 8 unequal squares at most.

And since 385-359 = 26 = 12 + 52, it is made of.

Remove 12 and 52, and the remaining 8 different squares are just enough.

22 + 32 + 42 + 62 + 72 + 82 + 92 +102 = 359, so 359 can be written as the sum of squares of up to 8 positive integers that are not equal to each other.

4) If the 6 numbers are different, then they must be 0 1 2 3 4 5

The numbers on 6 sheets of paper are marked A1-A6 on the front and B1-B6 on the reverse

Then A1+...a6-b1-..The value of b6 is |a1-b1|,.a6-b6|The algebraic sum of these values plus plus minus signs is 0

And there are odd numbers in 0 1 2 3 4 5, so they add symbols to the number of descendants, and the sum is an odd number, which cannot be 0

So it is impossible for 6 numbers to be different, i.e. there must be two numbers that are the same.

You have no solution to this equation, thank you.

If the coordinates of the intersection of the line y=-2x+a and the line y=2x+b are (m,10), then 10=-2m+a

10=2m+b ②

20=a+b

The image of a known primary function passes through the points (2,1) and (-1,-3) of the general formula of the primary function: y=ax+b

Substituting the coordinates of points (2,1) and (-1,-3) into the general formula.

1=2a+b

3=-a+b

Simultaneous solution of equations.

Get a = 4 3

b=-5/3

The expression is shown as y=4 3x-5 3

I'll do the second one, hee-hee.

Use the two-point formula (x-x1) (x-x2) = (y-y1) (y-y2) x-2) (x+1) = (y-1) (y+3).

You can do it.

The two upstairs mistakes are as follows:

The meaning of the first question is that at a certain point, the x-axis coordinates of the two straight lines are both m, and the y-axis coordinates are the same 10, so y=10=-2m+a, y=10=2m+b, and the formula + gives a+b=20

In the second question, because it is a one-time function, the expression is y=kx+b, and the two coordinates in the problem are brought into the expression, and 1=2k+b, -1=-3k+b, and the two formulas are combined to obtain k=2 5, b=1 5, so the expression of the primary function is y=2 5x+1 5

8A9, A, what are the conditions?

10. I can't see what's behind f(a).

7x17+8x18+9x19+10x20+71x7+81x8+19x9+20x10

Is your question wrong?

It should be 7x17+8x18+9x19+10x20+71x7+81x8+91x9+20x10, right?

If it is 7x17+8x18+9x19+10x20+71x7+81x8+91x9+20x10

10*20*2+19*9*2+7*17+71*7+18*8+81*820*20+19*18+7*88+8*9920*20+19*18+11*(7*8+8*9)I don't know.,I'm bad at math.。。。

(a+b) quadratic = a quadratic + 2ab + b quadratic = 13 + 12 = 25, so a + b = -5 or a + b = 5

a-b) quadratic = a quadratic - 2ac + b quadratic = 13 + 12 = 25, so a-b = 5 or a-b = 5

Find a+b because a square + b square = 13 ab=-6 so a square + b square + 2ab = (a + b) square = 13-12 = 1 so a + b = 1

Ask for a-b because.

a+b) squared = 1 ab=-6 so a-b squared = 1-(—24) = 25

It's too simple, isn't it? a²﹢b²＝13,ab＝﹣6 .then a b 2ab 13 2 6 1

The inverse application of the perfect square formula is simplification, and a b 2ab a b 1∴a＋b＝±√1＝±1. ﹙a－b﹚²＝a²﹢b²－2ab＝13－2×﹙﹣6﹜＝25.

a－b＝±√25＝±5.Answer: a b 1, a b 5

With a perfect sum of squares (difference) is fine.