Make a square iron sheet with an area of 108 square centimeters into a square box without cover. What is the maximum surface area of the box? The answer in the book is 67.5 square centimeters, but I made 60 square centimeters, A little low, hehe) A great Xia answered me like this (but I can't understand it, I hope the more detailed the better): you can draw along the diagonal of the square, and then there will be four small triangles and four large triangles. Four small triangles can be seen as a small square, and four large triangles can be seen as two small square Therefore, we can set the area of each small square as X 5X＋X＋2X＝108 X＝13.5 So its surface area is: S＝5X＝13.5×5＝67.5

Make a square iron sheet with an area of 108 square centimeters into a square box without cover. What is the maximum surface area of the box? The answer in the book is 67.5 square centimeters, but I made 60 square centimeters, A little low, hehe) A great Xia answered me like this (but I can't understand it, I hope the more detailed the better): you can draw along the diagonal of the square, and then there will be four small triangles and four large triangles. Four small triangles can be seen as a small square, and four large triangles can be seen as two small square Therefore, we can set the area of each small square as X 5X＋X＋2X＝108 X＝13.5 So its surface area is: S＝5X＝13.5×5＝67.5

To put it simply, if you cut off the four corners of a square, you will cut off four small triangles. Then you will cut out one big triangle on each of the four sides. The rest is just the shape you need to make a box. The area of the eight triangles cut out is three times that of the remaining square

If all the solutions of the equation LG (AX) LG (AX ^ 2) = 9 about X are greater than 1, the value range of a is obtained

lg(ax)*lg(ax^2)=4
(lga+lgx)*(lga+lgx^2)=4
If there is a question, a > 0
(lga+lgx)*(lga+2lgx)=4
2(lgx)^2+3lga*lgx+(lga)^2-4=0
Another lgx = t
Because LG (AX) * LG (AX ^ 2) = 4, all solutions are greater than one
So x > 1, so t = lgx > LG1 = 0
That is, all roots of the equation 2T ^ 2 + 3lga * t + (LGA) ^ 2-4 = 0 are greater than 1
therefore
a>0
-3lga/2>0
[(lga)^2-4]/2>0
Discriminant 9 (LGA) ^ 2-8 (LGA) ^ 2 + 32 > 0
The range of a satisfying the above four inequalities is 0

Given that a is the opposite of - 2012, B is the opposite of 7, and C is the number whose absolute value is equal to 3, find the value of a + b-c

a=-(-2012)=2012
b=-7
C = 3 or C = - 3
a+b-c=2012-7-3=2002
Or a + B-C = 2012-7 + 3 = 2008

Let f (x) = INX / X-X. (1) find the monotone interval of F (x) (2) let m > 0 find the maximum value of F (x) on [M, 2m]
2013 Hubei eight schools joint examination

(1) The definition field of F (x) is (0, + ∞)
　　f'(x)=(1-lnx)/x²-1=(1-lnx-x²)/x²
Let g (x) = 1-lnx-x & # 178;
　　g‘(x)=-1/x-2x

When a and B satisfy what conditions, the absolute value of a plus B = the absolute value of a plus the absolute value of B

a. B is greater than or equal to 0

Given the function f (x) = ln (1 + x) - x, if x is greater than 1, it is proved that (1-1 / (x + 1)) is less than or equal to (LN (1 + x)) is less than or equal to X

It seems that x > 0 is OK
Let g (x) = ln (1 + x) + 1 / (1 + x) - 1
f'(x)=1/(1+x)—1=—x/(1+x)

Derivative of absolute value of x minus 1
Notice the derivative of the absolute value of x minus 1. Thank you

When x > 1, it is less than 1

Given the function f (x) = a ^ 2X-4 + n (a > 0 and a ≠ 1), then M + n =?

A:
F (x) = a ^ (2X-4) + n constant crossing point P (m, 2)
That is, when 2X-4 = 0, x = 2:
f(2)=a^(2*2-4)+n=1+n=2
So: M = 2, n = 1
So: M + n = 3

It is known that the fourth power of polynomial (M + 2) x - the nth power + 7 of X is the third power, and the nth power of (m-1) can be obtained
I know it's very simple, but I don't want to think about it. I'll add points soon
It is known that the fourth power of polynomial (M + 2) x - the nth power of X + 7 is a binomial of degree 3, and the nth power of (m-1) can be obtained

If it is a cubic binomial, the maximum is three
So there's no x ^ 4
Then its coefficient is 0
So m + 2 = 0
m=-2
m-1=-1
The rest is up to three times
So n = 3
So the nth power of (m-1) = (- 1) & sup3; = - 1

The position of rational number ABC on the number axis is shown in the figure
————c——0——b————a————
Simplification: 1) - 3|c| + 4|b| + 2|b-a|
2)|c-a|+|b-c|+|a-c|

1)c0,(b-a)0
=a-c+b-c+a-c=2a+b-3c