The school library bought 12 bags of books, 4 books in each bag. Each book costs 25 yuan. How much did it cost to buy these books?

The school library bought 12 bags of books, 4 books in each bag. Each book costs 25 yuan. How much did it cost to buy these books?

25 × 4 × 12 = 100 × 12, = 1200 yuan. A: it costs 1200 yuan to buy these books

Mr. Kang and Mr. Li have the same money. Mr. Kang spends 58 yuan and Mr. Li spends 14 yuan. The amount of money left by Mr. Li is three times that of Mr. Kang
How many yuan did Mr. Li have? How many yuan do you need

3(X-58)=X-14
3X-174=X-14
2X=160
X=80

Solving inequality 12x ^ 2-ax-a ^ 2

The original formula = (4x + a) (3x-a) is: x = - A / 4 or x = 3 / A
Classification discussion: when A3 / A, so x > - A / 4 or x0, - A / 43 / A or X

Let f (x) have continuous derivatives on [0,1] and f (1) = 2, ∫ f (x) DX (1,0) = 3, then ∫ XF '(x) DX (1,0) =?
(1,0) after DX means that the upper limit of ∫ is 1 and the lower limit is 0

∫xf'(x)dx(1,0)
=∫(0,1)xdf(x)
=xf(x)|(0,1)-∫(0,1)f(x)dx
=f(1)-∫(0,1)f(x)dx
=2-3
=-1
Remember the integral limit, the lower limit is before, the upper limit is after, it should be (0,1), and put after the ∫ sign

"If a matrix A, B, C has ab = AC, then B = C" is that right?

It's not right
Only when a is full rank can we have this conclusion!

U = arctan (X-Y)

(in the figure above, the bottom line of output is followed by the answer.)

The derivative y '> 0 of function y = f (x) is monotonically increasing% e of function f (x)
Such as the title

It is a necessary but not a sufficient condition
If f '(x) > 0, the monotonic increase of F (x) cannot be deduced
Counter example: F (x) = - 1 / x, f '(x) = 1 / (x ^ 2) > 0 is constant, but f (x) does not increase monotonically
If f (x) increases monotonically, we can deduce that f '(x) > 0

For a fraction, the sum of the numerator and denominator is 63. If the denominator is added with 17, this number will get one third. What is the original fraction?

The numerator + denominator = 63
Denominator + 17 = 3 × numerator
The numerator is 20 and the denominator is 43
So the original score is 20 / 43

Please be able to solve the problem of the master solution of the following high number of differential equations, see when I am right or the answer is right
Li Yongle Wang Shian 2012 postgraduate entrance examination mathematics 2 p246. Question 6, 14, 15
The first line of questions, the second line of my answers, the third line of correct answers, and the fourth line of my methods may also be wrong
Maybe if I don't sit down and calculate with paper and pen, I can't say it out of thin air. I just hope you can check it for me and see if the answer is wrong!
Well, Feng on the third floor is right. I'm right about the first question. The answer to the second question is right. But I haven't found a way to solve the third question. Let's go tomorrow

1)ln(y)=ln[(x)/(x-4)]/4+C1
y=C2(x/(x-4))^(1/4)
So y ^ 4 = C (x) / (x-4)
You should be right
2) My results:
1 / (x * (1 / (2 * x ^ 2) + 1 / 2)) = (2 * x) / (x ^ 2 + 1) is the same as the correct answer
3) Using total differential equation to do

(2mn)²+（m²-n²）²

Original formula = 4m & # 178; n & # 178; + m ^ 4-2m & # 178; n & # 178; + n ^ 4
=m^4+2m²n²+n^4
=(m²+n²)²