If the value of the algebraic formula (2x to the second power + ax-y + 6) - (2bx to the second power - 2x + 5y-1) has nothing to do with the value of the letter X, find the algebraic formula ½ the second power of a - the second power of 2B + 4AB

If the value of the algebraic formula (2x to the second power + ax-y + 6) - (2bx to the second power - 2x + 5y-1) has nothing to do with the value of the letter X, find the algebraic formula ½ the second power of a - the second power of 2B + 4AB

solution
(2x²+ax-y+6)-(2bx²-2x+5y-1)
=(2-2b)x²+(a+2)x-6y+7
The value of ∵ is independent of X
∴2-2b=0
a+2=0
∴b=1,a=-2
∴1/2a²-2b²+4ab
=1/2×4-2+4×1×(-2)
=2-2-8
=-8

7x➗3=8.19

Two fifths of a meter, it can not only represent () of one meter, but also represent () of two meters

Two fifths of a meter, it can represent (2) fifths of a meter,
It can also denote (1 / 5) of 2 meters

If Sn = 48, s2n = 60, then s3n=______ ．

∵ sequence {an} is an equal ratio sequence, and its first n terms and Sn, s2n Sn, s3n-s2n also form an equal ratio sequence. ∵ (60-48) 2 = 48 × (s3n-60), the solution is s3n = 63, so the answer is 63

The second race starts from the level road and starts downhill 4 km from the midpoint. After driving through the midpoint for 26 km, it is all uphill and downhill
The second race starts from the level road and starts downhill 4 km from the midpoint. After driving through the midpoint for 26 km, it is all uphill
It is known that a car takes the same time in the two races; the speed of the second race is 5 / 6 of the speed of the first race, and the speed will decrease by 25% when encountering an uphill and increase by 25% when encountering a downhill, What is the distance of each race?
To solve the equation!

The first race, Pinglu 6 uphill speed 6 × (1-25%) = 9 / 2 downhill speed 9 / 2 × (1 + 25%) = 45 / 8, the second race, Pinglu 5 downhill speed 5 × (1 + 25%) = 25 / 4 uphill speed 25 / 4 × (1-25%) = 75 / 16, draw a picture, the distance around the midpoint is 22 + 4 + 4 + 22 km, the remaining distance is equal to the first race, uphill 2

The section of the reservoir dam is trapezoidal ABCD, the crest ad = 6m, the slope length CD = 5 pieces of 6m, the bottom BC = 21 + 5 pieces of 3M angle ADC = 135 ° and the slope angle ABC is calculated

For de ⊥ BC, AF ⊥ BC, e and f respectively, 〈 ADC = 135 °, then 〈 C = 45 °, de = dcsin, 45 ° = 8 * {2 / 2 = 4} 2, CE = de = 4} 2, Fe = ad = 6, BF = bc-ce-ef = 30-6-4} 2 = 24-4} 2, AF = de = 4} 2, tanb = AF / BF = 4} 2 / (24-4} 2) = (3} 2 + 1) / 17 ≈ 0.30835, 〈 B ≈ 17.1

A math problem
After six days of building a highway, it is more than 120 meters above the midpoint, and the rest will continue to be built at the original speed. It can be completed in five years. How many meters of this highway?
The teacher's answer is 1320 meters, mine is 2640 meters

2640m
6v=1/2s+120
s=11v
v=240 s=2640

It is known that the matrix A of order n satisfies a ^ 2 = a + 6I. It is proved that 1). The determinant of a is not equal to 5 2). When the determinant of a = 72, find n
It is known that the matrix A of order n satisfies a ^ 2 = a + 6I. It is proved that 1). The determinant of a is not equal to 5 2). When the determinant of a = 72, find n

1) If a matrix of order n satisfies a ^ 2 = a + 6I, let X be any eigenvalue of a and a be the eigenvector of X, then
Aa=xa
From a ^ 2 = a + 6I, a ^ 2-a-6i = 0
So (a ^ 2-a-6i) a = 0,
A^2a-Aa-6a=0,
x^a-xa-6a=0
Because a ≠ 0
So x ^ 2-x-6 = 0
X = 3, or x = - 2
The determinant of a is equal to the product of all its eigenvalues, so the determinant of a is not equal to 5
2) It should be when the determinant of a = - 72, because - 72 = 3 × 3 × (- 2) × (- 2) × (- 2), which is the product of five eigenvalues
n=5.

Fill in the box with nine numbers 1 ` 2 ` 3 ` 4 ` 5 ` 6 ` 7 ` 8 ` 9 to establish the formula "each number shall not be used repeatedly"

2 9 4
7 5 3
6 1 8

Let a be a square matrix of order n, B be an n × s matrix, and R (b) = n. It is proved that if AB = 0, then a = 0

If AB = 0, then the column vectors of B are solutions of AX = 0
Because R (b) = n, ax = 0 has at least n linearly independent solutions
Let the solution set be s, then R (s) = N-R (a) > = n
That is R (a) = 0
So r (a) = 0
That is, a = 0