Fill - 1, + 2, - 3, + 4, - 5, + 6, - 7, + 8, - 9 in the grid of the right graph, so that the three numbers on each row, column and diagonal satisfy the following two conditions: (1) the product of the three numbers is negative; (2) the sum of the absolute values of the three numbers is equal

Fill - 1, + 2, - 3, + 4, - 5, + 6, - 7, + 8, - 9 in the grid of the right graph, so that the three numbers on each row, column and diagonal satisfy the following two conditions: (1) the product of the three numbers is negative; (2) the sum of the absolute values of the three numbers is equal

As shown in the figure: the answer to this question is not unique

A primitive function SiNx of F (x), then the integral of F (x) ∧ n

If the original function of F (x) is SiNx, that is, the derivative of SiNx is f (x).. then f (x) = cosx
The integral of F (x) ^ n = (cosx) ^ n, (cosx) ^ n can be found in the appendix table of Tongji edition of higher mathematics. I can't remember it here. Sorry

One variable linear equation of absolute value
It is known that the power of (1-m absolute value) times X - (M + 1) x + 8 = 0 is a linear equation of one variable about X, and the power of 2 (M + n) absolute value + (2P + VN) is 0, so the value of 1 / 2np can be obtained

(absolute value of 1-m) = 0, M = plus or minus 1
(M + 1) is not equal to 0, so m = 1, minus one is rounded off
The absolute value of 2 (M + n) + (2P + VN) = 0
So the absolute value of 2 (M + n) is 0 and the square of + (2P + VN) is 0
So n = negative 1,2p + VN = 0
So 2p = - VN = V, P = V / 2
1 / 2np = negative 1 / V

Given that f (x) has f (x + y) = FX + FY for all real numbers x, y belonging to R, and when x > 0, FX < 0, F3 = - 2, try to judge the parity of function on R

From F (3) = - 2, f (3) = f (1.5 + 1.5) = 2F (1.5) f (1.5) = - 1F (1.5) = f (3-1.5) = f (3) + F (- 1.5) = - 1, then f (- 1.5) = 1F (0) = f (1.5-1.5) = f (1.5) + F (- 1.5) = 0, so f (0) = f (x-x) = f (x) + F (- x) = 0f (x) = - f (- x) is an odd function

Using Taylor formula to find limit when x tends to infinity [x-x ^ 2ln (1 + 1 / x)]

Let t = 1 / X
Original formula = Lim [T-ln (1 + T)] / T ^ 2 T - > 0
ln(1+t)=t-t^2/2+o(t^2)
So the original formula = Lim [T-T + T ^ 2 / 2] / T ^ 2 = 1 / 2 + O (1) = 1 / 2

Let a > 0, f (x) = e ^ X / A + A / e ^ X be the even function on R. (1) find the value of a (2) prove that f (x) is an increasing function on (0, + ∞) (3) solve the equation f (x) = 2

A ＞ 0, f (x) = e ^ X / A + A / e ^ x is an even function on R,
That is, e ^ X / A + A / e ^ x = e ^ (- x) / A + A / e ^ (- x),
∴(e^x-1/e^x)(a-1/a)=0,
∴a-1/a=0,a^2=1,a>0,
∴a=1.
(2)f(x)=e^x+e^(-x),x>0,
f'(x)=e^x-e^(-x)>0,
F (x) is an increasing function
（3）f(x)=2,
(e^x)^2+1=2e^x,
(e^x-1)^2=0,
e^x=1,x=0.

Let the real symmetric matrix a satisfy (A-E) (A & # 178; + e) = 0, and prove that a = E

Because (A-E) (A & # 178; + e) = 0
So the eigenvalue a of a satisfies (A-1) (a ^ 2 + 1) = 0
Because the eigenvalues of real symmetric matrix are real numbers
So a = 1
So the eigenvalues of a are 1,1,..., 1
And because the real symmetric matrix can be diagonalized
So a = pdiag (1,1,..., 1) P ^ - 1 = PEP ^ - 1 = E

Why does have become has when it is plural? Isn't have singular

Have is the original form of the word and has is its third person singular form
In the simple present tense, when the subject is the third person singular, the predicate verb uses has
She has a book.
I have a book.
They have a lot of books.

Simplification brings in evaluation 1 / 2 (- 3ax & # 178; - ax + 3) - (- ax & # 178; - 1 / 2ax-1) (where a = - 2x = 3)

1/2（-3ax²-ax+3)-(-ax²-1/2ax-1)
=-3/2ax²-1/2ax+3/2+ax²+1/2ax+1
=-1/2ax²+5/2
Substituting - 1 / 2 * (- 2) * 3 & # 178; + 5 / 2
=9+5/2
=23/2

Are "plenty of", "a great deal of" and "a lot of" followed by countable nouns or uncountable nouns?

Can only modify uncountable nouns: much, (a) little, quite a little, a bit of, amount of, a (good / great) deal of