After dividing the - polynomial [(17x2-3x + 4) - (AX2 + BX + C)] by (5x + 6), the quotient is (2x + 1), and the remainder is 0. Find a-b-c = () A. 3B. 23C. 25D. 29

After dividing the - polynomial [(17x2-3x + 4) - (AX2 + BX + C)] by (5x + 6), the quotient is (2x + 1), and the remainder is 0. Find a-b-c = () A. 3B. 23C. 25D. 29

According to the meaning of the title, we get [(17x2-3x + 4) - (AX2 + BX + C)] = (5x + 6) (2x + 1), (17-a) x2 + (- 3-B) x + (4-C) = 10x2 + 17x + 6, 17-a = 10, - 3-B = 17, 4-C = 6, the solution is a = 7, B = - 20, C = - 2, a-b-c = 7 + 20 + 2 = 29

After dividing a polynomial (3x2-x + 8) - (AX2 BX + C) by 2x, the quotient (3x + 1) is obtained. The remainder is 0. The value of a-b-c is obtained

(3x2-x+8)-(ax2-bx+c)=2x(3x+1)
∴(3-a)x²+(b-1)x+(8-c)=6x²+2x
∴3-a=6
b-1=2
8-c=0
∴a=-3
b=3
c=8
∴a-b-c
=-3-3-8
=-14

Given the function f (x) = LG (1-x) + LG (1 + x) + x ^ 4-2x ^ 2, find its range

∵ 1-x ＞ 0 1 + X ＞ 0 - 1 ＜ x ＜ 1} domain: (- 1,1) f (- x) = LG ((1 + x) + LG (1-x) + - x) ^ 4-2 (- x) ^ 2 = f (x)} function f (x) is even function. F (x) = LG (1-x) + LG (1 + x) + x ^ 4-2x ^ = LG ((1-x) ((1 + x)) + x ^ 2 (x ^ - 2) = LG (1-x) + x ^ 2 (x ^ 2-2) ∵ - 1 ＜ x ＜ 1

The edge length of a small square is 4cm, and that of a large square is 8cm. The edge length ratio of a small square to a large square is [], the surface area ratio is [], and the volume ratio is

The edge length ratio of small square and large square is [1:2], the surface area ratio is [1:4], and the volume ratio is (1:8)

How can 3 11 4 10 be equal to 24

lg10=1
(4-lg10)x(11-3)=24

If the sum of the first n terms of the sequence {an} is Sn, there is 6sn = 1-2an 1 for any positive integer n, the general term formula of the sequence {an} is obtained
If the sum of the first n terms of the sequence {an} is Sn, 6sn = 1-2an for any positive integer n
1, find the general term formula of sequence {an}!

Sn=(1-2an)/6
an=Sn-S(n-1)
=(1-2an)/6-[(1-2a(n-1)]/6
an/a(n-1)=1/4
So {an} is an equal ratio sequence with a common ratio of 4
a1=S1=(1-2a1)/6
a1=1/8
an=1/8*1/4^(n-1)
=1/2^(2n+1)

It is known that the perimeter of a large square is 48CM longer than that of a small square, and their area difference is 480cm2
If the side length of a large square is ACM and that of a small square is BCM, then a-b=
Please find out the side length of two squares

The solution is that the perimeter of the large square is 48CM longer than that of the small square
Then 4a-4b = 48
That is, A-B = 12
And by the difference of their area
Then a ^ 2-B ^ 2 = 480
That is, (a + b) (a-b) = 480
That is, (a + b) 12 = 480
That is, a + B = 40. ②
From (1) (2) simultaneous solution
a=26,b=14

Can we simplify the process of (4 and 7 / 4-1 and 14 / 11) / (7 / 12 + 5 / 28) * 8 / 13

Original formula = (4 + 4 / 7-1-11 / 14) / [(49 + 15) / 84] × 8 / 13
=（3-3/14）/（16/21）×8/13
=39/14×21/16×8/13
=（39×21×8）/（14×16×13）
=（9×13×7×8）/（4×7×8×13）
=9/4

For any real number x1, X2, Max {x1, X2}, denote the larger number in x1, X2, then when x belongs to R, the function f (x) = max {2-x & # 710; 2,2},
What is the difference between the maximum and minimum of X belonging to [- 3,1 / 2]

∵ real numbers x1, X2, Max {x1, X2} denote the larger number of x1, x2,
∵x∈[-3,],
When x = 0, the maximum value of 2-x2 is 2,
∴f（x）=max{2-x2,x}=2,
So the answer is 2

Lay a square with 15 cm long and 8 cm wide rectangular tiles. How many cm is the minimum side length of this square? How many such tiles do you need at least?

Using a rectangular tile 15 cm long and 8 cm wide to lay a square is the problem of finding the least common multiple of 15 and 8
The least common multiple of 15 and 8 is 120
So the side length of a square is 120 cm
At least how many such tiles (120 △ 15) × (120 △ 8) = 120