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Vector / A + B / = 1, B = (3,4), when / A / reaches the maximum, vector a =?
|In this case, a = (- 3 / 5, - 4 / 5)
What is the condition that a = B is the absolute value a = B?
A = B is a sufficient and unnecessary condition for absolute value a = absolute value B,
The absolute value of a is equal to
a,-a,0
For any real number x, | x + 1 | + | X-2 | > A is constant, and the value range of a is obtained
∵|x + 1 | + |x-2 | = |x + 1 | + |2-x | ≥ |x + 1 + 2-x | = 3, the minimum value of |x + 1 | + |x-2 |, is 3, the range of a is (- ∞, 3)
The absolute value of any real number x, X-1 - x + 3 ＜ A is constant, and the value range of a is constant
|x-1|-|x+3|
If there is a real number x such that the absolute value sign x-a, absolute value sign + absolute X-1, absolute ≤ 3 holds, then the value range of the real number a
It can be concluded from the questions:
|x-a|+|x-1|≤3
|X - a | can be regarded as the distance from the number axis to the point represented by A;
|X - 1 | can be regarded as the distance from the number axis to the point represented by 1;
Make the sum of the two distances less than or equal to 3
That is to say, if the sum of two line segments does not exceed 3, then it is exactly 3, a = 4 or a = - 2
Then there is a real number such that the inequality holds when - 2 ≤ a ≤ 4
Therefore, the value range of a is - 2 ≤ a ≤ 4
Given that the value of square + 3 of algebraic formula A is 3, then the value of square + 6a-5 of algebraic formula 2a is?
It should be a & # 178; + 3A = 3
So the original formula = 2 (A & # 178; + 3a) - 5
=2×3-5
=1
Given a plus B = 6, find the square of the algebraic formula a [a plus b] - a plus B 6A
Brothers and sisters, help quickly. Thank you very much
(a + b) / A & # 178; = 6, so a & # 178; / (a + b) = 1 / 6,
[3(a+b)]/a² - (6a²)/(a+b)
= 3×6 - 6×(1/6)
=18-1 = 17
It is known that the absolute value of a + 3 plus the absolute value of B-2 plus the absolute value of C + 5 is equal to 0
/a+3/+/b-2/+/c+5/=0
=/0/+/0/+/0/=0
=/-3+3/+/2-2/+/-5+5/=0
=a=-3 b=2 c=-5
Given that the absolute value of a is equal to 3, B = (1,2), and a is perpendicular to B, the coordinates of a are obtained
Let the coordinates of vector a = (x, y) ∵ vector b = (1,2), and vector a be perpendicular to vector B ∵ x + 2Y = 0 (1) The absolute value of vector a is equal to 3, the square of X + the square of y = 9 (2) Simultaneous equations (1) and (2), the solution of the equations: x = 6 √ 5 / 5, y = - 3 √
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