A proof of differential mean value theorem It is proved that the equation x ^ 5 + X-1 = 0 has only one positive root

A proof of differential mean value theorem It is proved that the equation x ^ 5 + X-1 = 0 has only one positive root

You asked about the exercises of the fifth edition of advanced mathematics in Tongji University
If you really want to understand, I suggest you ask the professor, after all, the Internet is not clear
The mean value theorem is to find the extremum of a function, which increases monotonically and has no extremum
Because f '(x) = 5x ^ 4 + 1 is always greater than zero (does not satisfy the theorem), there is no extremum in R, and f (0) = - 1 is less than 0, so the function has a positive root

It is known that the parabola y = ax ^ 2-2x + C intersects its axis of symmetry at point a (1, - 4), at point C with the Y axis, and at point B with the positive half axis of X axis
(1) Find the function relation of this parabola;
(2) Let a straight line AC intersect X-axis at point D, p be a moving point on line ad (point P is different from points a and D), cross a straight line AB at point e through P as PE / / x-axis, and cross e as EF ⊥ X-axis at point F, then calculate the coordinates of point P when the area of quadrilateral oPEf is equal to 7 / 2

(1) Because only the vertex of the parabola is on the axis of symmetry, so a is the vertex of the parabola, so the axis of symmetry is x = - (- 2) / (2a) = 1 / a = 1, so a = 1 let the analytical formula of the parabola y = x & # 178; - 2x + C be substituted into the point (1, - 4) 1-2 + C = - 4, C = - 3. The analytical formula of the parabola is y = x & # 178; - 2x-3 (2) C is the intersection of the throwing line and the Y axis, so C

Factorization of 31 × 3.14 + 27 × 3.14 + 42 × 3.14
emergency

31×3.14+27×3.14+42×3.14
=(31+27+42)×3.14
=100×3.14
=314

It is known that the triangle ABC is not a right triangle, the angle a = 50 degrees, BD and CE are high, and their lines intersect at the H point. The degree of the angle BHC is calculated

50 degrees
It is known that CE is perpendicular to AB, so the angle AEC is equal to 90 degrees
Similarly, the angle ADC is equal to 90 degrees
Angle a + angle B + angle c + angle D equals 360 degrees
Because a equals 50
So 50 + 90 + 90 + angle EHD equals 360 degrees
So the angle dne is 50 degrees
Because angle a is equal to angle BHC
So ∠ BHC = 50 degrees

A simple algorithm of 75 × 27 + 19 × 25

75×27＋19×25
=(25×3)×27＋19×25
=25×(3×27)＋19×25
=25×81＋19×25
=25×(81＋19)
=2500

How to change determinant into triangle determinant
Is there any general way to give a determinant? I don't know how to change it for a long time

To calculate determinant with property trigonometry, we usually deal with it from left to right column by column. First, we exchange a relatively simple (or small) non-zero number to the upper left corner (in fact, it is also changed to the last row), and use this number to eliminate the rest of the number in the first column to zero

384 × 3 / 4-3 / 8 divided by 4 / 3
2.2 × 0.75 + 2.2 × 1 / 4
Seventy two divided by five sixths + five eighths minus one third
The sum of four fifths divided by 32 × 1 minus five fifths and plus three fifths

384×（3/4-3/8）÷4/3
=384×3/8÷4/3
=144×3/4
=108
2.2×0.75+2.2×1/4
=2.2×（0.75+1/4）
=2.2×1
=2.2
72÷（5/6+5/8-1/3）
=72÷9/8
=64
4/5÷32×（1-5/8+3/5）
=1/40×39/40
=39/1600

Given the proposition p: "for any x belongs to [1,2], x ^ 2-A > = 0", proposition q: "exist, x0 belongs to R, x0 ^ 2 + 2ax0 + 2-A = 0", if the proposition "P and Q have and only one true proposition, find the value range of real number a"

From proposition p, we can get: X & # 178; ", a, and X ∈ [1,2], then a" 1 "can be satisfied. For proposition q, as long as △ = B & # 178; - 4ac" 0, we can get a "- 1 or a" 0, so 0 "a" 1 or a "- 1
If you don't understand, please take it

Simple calculation of 5 * 6 * 7 * 8-2 * 3 * 4 * 5 out of 5 * 6 * 7 * 8

(5*6*7*8-2*3*4*5)/5*6*7*8
=5*6*7*8/5*6*7*8-2*3*4*5/5*6*7*8
=1-1/14
=13/14

1/2x+3(-2z+y)-(1/3x-4y-5z)

1/2x+3(-2z+y)-(1/3x-4y-5z)
=(1/2)x-6z+3y-(1/3)x+4y+5z
=(1/2-1/3)x+(-6+5)z+(3+4)y
=(1/6)x-z+7y