If two tangents of circle x + y-2ax + A + 2a-3 = 0 are made through point a (a, a), what is the value range of real number a?

If two tangents of circle x + y-2ax + A + 2a-3 = 0 are made through point a (a, a), what is the value range of real number a?

X ^ 2 + y ^ 2-2ax + A ^ 2 + 2a-3 = 0 (x-a) + y = 3-2a > 0 a3-2a a + 2a-3 > 0 (a + 3) (A-1) > 0 a > 1 or a
It is known that X and y are rational numbers. Now a new operation method "*" is provided, which satisfies the condition of X * y = XY + 1 and tries to find the value of 2 * 4
2※4=2*4+1=9
Simplify the following formulas 8x - (- 3x-5) (3x-1) - (2-5x) (- 4Y + 3) - (- 5y-2) 3x + 1-2 (4-x)
8x-(-3x-5)
=8x+3x+5
=11x+5
(3x-1)-(2-5x)
=3x-1-2+5x
=8x-3
(-4y+3)-(-5y-2)
=-4y+3+5y+2
=y+5
3x+1-2(4-x)
=3x+1-8+2x
=5x-7
Use the square difference formula to calculate: 19 7 / 9 * 20 2 / 9
19 and 7 / 9 * 20 and 2 / 9
=(20-2/9)*(20+2/9)
=20^2-(2/9)^2
=400-4/81
=399 and 77 / 81
If passing through point a (a, a) can make two tangents of circle x ^ 2 + y ^ 2-2ax + A ^ 2 + a-6 = 0, then the range of real number a is
x^2+y^2-2ax+a^2+a-6=0
(x-a) ² + Y & #178; = 6-A, 6-A > 0 a radius
|A | > under radical (6-A)
a²>6-a
a²+a-6>0
(a+3)(a-2)>0
-3
It is known that X and y are rational numbers. Now a new operation ⁃ is provided, which satisfies x ⁃ y = XY + 1. (1) find the value of 2 ⁃ 4; (2) find the value of (1 ⁃ 4) ⁃ (- 2); (3) select any two rational numbers (at least one is negative), fill in the following ⁃ and ⁃ respectively, and compare their operation results: ⁃ and ⁃ 0 ⁃; (4) explore the relationship between a ⁃ B + a ⁃ C, And express them with equations
（1）2※4=2×4+1=9；（2）（1※4）※（-2）=（1×4+1）×（-2）+1=-9；（3）（-1）※5=-1×5+1=-4，5※（-1）=5×（-1）+1=-4；（4）∵a※（b+c）=a（b+c）+1=ab+ac+1，a※b+a※c=ab+1+ac+1．∴a※（b+c）+1=a※b+a...
3x+4y=7 2x+y+3z=9 5x+7y-9y=8
3x+4y=7 （1）
2x+y+3z=9 （2）
5x+7y-9y=8（3）
（2）×3+（3）
11x+10y=35（4）
3x+4y=7（1）
（1）×11-（4）×3
44y-30y=77-105
14y=-28
Y = - 2 substituting (1)
3x-8=7
3x=15
X=5
Substituting (2)
10-2+3z=9
3z=1
z=1/3
therefore
X=5
y=-2
z=1/3
Using the square difference formula to calculate 19 2 / 3 * 20 1 / 3
Use square difference formula to calculate 2008 ^ 3-2007 * 2008 * 2009 use square difference formula to calculate (1-1 / 2 ^ 2) (1-1 / 3 ^ 2) (1-1 / 4 ^ 2) （1-1/10^)
Calculate 2009 ^ 2-2 * 2009 * 9 + 9 ^ 2 with complete square formula
2008^3-2007*2008*2009
= 2008^3-(2008-1 )*2008*(2008+1)
Simplification
=20008^3-2008^3+2008^2
=2008^2
2009^2-2*2009*9+9^2
=2009^2-2*2009*9+9*9
=2009^2+9*9-2*2009*9
=（2009-9）^2
Original formula = 2008 & # 179; - (2008-1) (2008 + 1) 2008
=2008³-2008³+2008
=2008
Original formula = (1 + 1 / 2) (1-1 / 2) (1 + 1 / 3) (1-1 / 3) (1 + 1 / 4) (1-1 / 4) · (1 + 1 / 10) (1-1 / 10)
=... unfold
Original formula = 2008 & # 179; - (2008-1) (2008 + 1) 2008
=2008³-2008³+2008
=2008
Original formula = (1 + 1 / 2) (1-1 / 2) (1 + 1 / 3) (1-1 / 3) (1 + 1 / 4) (1-1 / 4) · (1 + 1 / 10) (1-1 / 10)
=1/2·3/2·2/3·4/3·3/4·4/5···9/10·11/10
=1/2·11/10
=11/20
The original formula = (2009-9) &# 178; = 4000000 question: using the square difference formula to calculate 19 2 / 3 * 20 1 / 3
A (a, a) can be used as two tangents of the circle x ^ 2 + y ^ 2-2ax + A ^ 2 + 2a-3 to show that point a is outside the circle, that is, substituting point a into x ^ 2 + y ^ 2-2ax + A ^ 2 + 2a-3
Our teacher talked about this question, but his answer seems to be 1.5 > a > 1 or a
(x-a)^2+y^2=-2a+3
Then - 2A + 3 > 0
a√(-2a+3)
square
a^2>-2a+3
a^2+2a-3=(a+3)(a-1)>0
A1
So a
Search the relationship between a * (B + C) and a * B + A * C, and express them by equation
Given that X and y are rational numbers, a new operation *, which satisfies x * y = XY + 1 (1) to find the value of 2 * (- 4); (2) [1 * 4] * (- 2); (3) to search the relationship between a * (b + C) and a * B + A * C, and express them by equation
（1）2*（-4）=-8+1=-7
（2）1*4=5
5*（-2）=-10+1=-9
（3）a*（b+c）=ab+ac+1
a*b+a*c=ab+1+ac+1
So a * (B + C) = a * B + A * C-1