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Anonymous users2024-02-11a represents an even set, b represents an odd set, and a u b = set of integers = z
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Anonymous users2024-02-10,-1,-2).(1,2,-1)=3
a*b=|i j k|=5i-j+7k
2a).3b=
Vectors can be represented by directed line segments. The length of a directed segment represents the size of the vector, and the size of the vector is the length of the vector. A vector of length 0 is called a zero vector.
Write as a vector whose length is equal to 1 unit.
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Anonymous users2024-02-09Today at 12:57 enthusiastic netizens fastest,-1,-2)(1,2,-1)=3
a*b=|i j k|=5i+j+7k
2a).3b=
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Anonymous users2024-02-08Because k belongs to z, 2k-1 is an odd number.
2k is an even number.
So aub = all integer z
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Anonymous users2024-02-07a=b=
x=2k+1=2(k+1)-1
Let b=k+1, then x=2b-1, k z, then b=k+1 z, so a=xianzen wantonly whispers that it is equal to b, in fact, the initiation with it is an odd set.
If you don't understand, please hi me and have fun studying!
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Anonymous users2024-02-06Let a=, b=, and verify: a=b
3.Solve the system of equations 2x+ y = 2y=8 c-2y=8d.
4v=8 solution:b [Analysis] The result of subtracting the two equations in the system of cherry closure equations can be used to determine the fracture of the ridge state 2x- 3y =2 .Detailed Sensitive Solution] Solve the system of equations bj, h0-0y-12x+ y =10
2(-39) = 10-2, that is, 49 = 8, so choose b
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Anonymous users2024-02-05a:2 5 8 11 14 17 20 23 26…Leadmin....b:2 8 14 20 26……
Apparently b is included in a
In fact, a can be written as 3x+2, and the meaning of the socks is the same as 3x-1.
Both 3x+2 and 6k+2 have 2s, the former with 3 plus 3 and the latter with 6 plus 6.
Affirmation that the former contains the latter, and that the latter contains the former!
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Anonymous users2024-02-04First question:
Use the method of rebuttal to the enlightenment.
Suppose 4k 2 a, then there are: a 2 b 2 4k 2 even numbers, a and b must be even numbers.
can make a 2m and b 2n, where m and n are integers. Gotta :
a^2－b^2＝4m^2－4n^2＝4（m^2－n^2）＝4k－2.
Obviously, 4 (m 2 n 2) can be divisible by 4 whole rounds, while 4k 2 is not divisible by 4, a 2 b 2 4k 2 cannot be true.
4K 2 is not part of the set A
Second question:
When a 2m, b 1 (and a 2m m is an integer), a 2 b 2 4m 2 1 2 (2m 2) 1
Let k 2m 2, get: a 2 b 2 2k 1
2k 1 set a
a^2b+b^2c+c^2a-ab^2-bc^2-ca^2a^2(b-c)+a(c^2-b^2)+bc(b-c)a^2(b-c)-(ab+ac)(b-c)+bc(b-c)(b-c)(a^2-ac-ab+bc) >>>More
sn+s（n-1）=an^2
s(n-1)-s(n-2)=a(n-1) 2 is subtracted by the formula, and the left side is an+an-1 >>>More
Hello! This question is a question to investigate the basic properties of inequalities, and the answer is as follows, 1 n 2+1 m 2=(1 n) 2+(1 m) 2>=2*(1 n)*(1 m), so 1 mn<=((1 n) 2+(1 m) 2) 2=(a 2+b 2) (2*a 2*b 2), multiply 1 2mn by 1 2 on the left and right, and get 1 2mn<=(a 2+b 2) (4*a 2*b 2), so the maximum value of 1 2mn is (a 2+b 2) (4*a 2*b 2), I wish you good progress!
Answer:- 6 b 4
1/2≤sinb≤ √2/2 >>>More
sin 2x-cos 2x=1 5 can be written as (sin 2x-cos 2x) 1=1 5
sin^2x-cos^2x）/1=1/5 >>>More