If the intersection of two straight lines y = x + 2a and y = 2x + A + 1 is p, and P is inside the circle x2 + y2 = 4, then the value range of a is______ .

If the intersection of two straight lines y = x + 2a and y = 2x + A + 1 is p, and P is inside the circle x2 + y2 = 4, then the value range of a is______ .

By solving the equations y = x + 2ay = 2x + A + 1, P (A-1, 3a-1), ∵ P is in the interior of the circle x2 + y2 = 4, ∵ Po | 2 < 4, that is: (A-1) 2 + (3a-1) 2 < 4 ∵ 15 < a < 1, so the value range of a is (- 15, 1)
It is known that three of the four numbers x + y, X-Y, XY and X / y have the same values. All pairs (x, y) with such properties are obtained
If x + y = X-Y, then y = 0, then x / y is meaningless, so x + y and X-Y are not equal. We can get x + y = xy = x / Y ≠ X-Y (1) or X-Y = xy = x / Y ≠ x + y (2) solution (1) gets y = - 1, x = 1 / 2 solution (2) gets y = - 1, x = - 1 / 2. Therefore, there are two groups of such number pairs (x, y)
The solution inequality of 4x of 2 minus 1 is greater than or equal to 1 of 6 minus 5x
(4x-1) / 2 ≥ (1-5x) / 6, remove the denominator, multiply both sides of the inequality by 6 at the same time
3 (4x-1) ≥ 1-5x, remove brackets:
12x-3 ≥ 1-5x
The combination of the same category is more than or equal to 12x + 1, and the same category is more than or equal to 5x
17x≥4
x≥4/17
(4x-1)/2≥(1-5x)/6
12x-3≥1-5x
17x≥4
x≥4/17
4x/2-1>=1/6-5x
Multiply the general score by 6 to get 12x-6 > = 1-30x
42x>=7
x>=1/6
Go to the denominator first and multiply the three terms by six to get 3 (4x-1) ≥ 1-30x 12x-3 ≥ 1-30x 42x ≥ 4 x ≥ 2 / 21
(4x-1)/2>(1-5x)/6;
3*(4x-1)>(1-5x);
12x-3>(1-5x);
17x>4;
The inequality solution is: x > 4 / 17
Simplify the steps of solving the problem x + 1 | + | X-2 | - | x-3 | with the method of zero interval discussion
The so-called zero points are x = - 1, x = 2 and x = 3, which divide the number axis into four parts, namely four intervals
① When x
If the intersection point P of two straight lines y = x + 2a, y = 2x + A is inside the circle (x-1) 2 + (Y-1) 2 = 4, then the value range of real number a is ()
A. - 15 ＜ a ＜ 1b. A ＞ 1 or a ＜ - 15C. - 15 ≤ a ＜ 1D. A ≥ 1 or a ≤ - 15
Simultaneous y = x + 2a, y = 2x + A, the solution is x = ay = 3A, the intersection point P (a, 3a) of two straight lines y = x + 2a, y = 2x + a. ∵ the intersection point P is in the interior of circle (x-1) 2 + (Y-1) 2 = 4, ∵ (A-1) 2 + (3a-1) 2 < 4, which is reduced to 5a2-4a-1 < 0, the solution is − 15 < a < 1. ∵ the value range of real number a is (− 15, 1). Therefore, select a
If the rational numbers a and B are opposite to each other, and X and y are reciprocal to each other, then the value of (a + b) x / y-xy is__________ .
Rational numbers a and B are opposite to each other, and X and y are reciprocal to each other
a+b=0
xy=1
therefore
The value of (a + b) x / y-xy is = - 1
Solving inequality [4x ^ 2-20x + 18 / x ^ 2-5x + 4] > 3
[4x & # 178; - 20x + 18] / [x & # 178; - 5x + 4] > 3 [transfer, general division, factorization]
On the number axis [- 2] / (x) - 1]
The solution set is {x | X
(x ^ 2-5x + 4) > 0, i.e. X4
There are 4x ^ 2-20x + 18 > 3 (x ^ 2-5x + 4)
That is, (X-2) (x-3) > 0 is X3
Common solution, X4
When (x ^ 2-5x + 4)
1 / 3 [1 / 4 (X-2) + 3] = 7 solutions
1 / 3 [1 / 4 (X-2) + 3] = 7 [multiply both sides by 3]
1 / 4 (X-2) + 3 = 21 [multiply both sides by 4]
x-2+12=84
x=84+2-12=74
X is 74
Given that the point P (1,1) is inside the circle (x-a) 2 + (y + a) 2 = 4, then the value range of real number a is______ .
∵ point P (1,1) is in the interior of the circle (x-a) 2 + (y + a) 2 = 4, ∵ (1-A) 2 + (1 + a) 2 ＜ 4. That is, A2 ＜ 1. The solution is: - 1 ＜ a ＜ 1. The value range of real number a is (- 1,1). So the answer is: (- 1,1)
The value of (X & sup2; - 4) (X & sup2; - 10x + 21) + 100 must be nonnegative by using the knowledge of integral and factorization
(x^2-4)(x^2-10x+21)+100
=(x^2-4)*[(x-5)^2-4]+100
=X ^ 2 * (X-5) ^ 2-4 * (X-5) ^ 2-4 * (x ^ 2-4) + 16 + 100
=X ^ 2 * (X-5) ^ 2-8 * x * (X-5) + 16 is the key step
=[x*(x-5)-4]^2 >=0
The value of (X & # 178; - 4) (X & # 178; - 10x + 21) + 100 must be non negative
Original formula = (x ^ 2-4) * [(X-5) ^ 2-4] + 100
=X ^ 2 * (X-5) ^ 2-4 * (X-5) ^ 2-4 * (x ^ 2-4) + 16 + 100
=x^2*(x-5)^2-8*x*(x-5)+16
=[x*(x-5)-4]^2≥0
When the power supply is not enough in the system, please handle the complex volume of the mixing tank from 1 to 6 months
(x²-4)(x²-10x+21)+100
=(x²-4)×[(x-5)²-4]+100
=X & # 178; (X-5) & # 178; - 4 (X-5) & # 178; - 4 (X & # 178; - 4) + 16 + 100
=X & # 178; (X-5) &# 178; - 8x (X-5) + 16 is the key step
=[x(x-5)-4]²≥0
The value of (X & # 178; - 4) (X & # 178; - 10x + 21) + 100 must be non negative