It is known that the power of 3 with 5 as the base of log is a, the power of 4 with 5 as the base of log is 4, and the power of 12 with 25 as the base of log is represented by A.B

It is known that the power of 3 with 5 as the base of log is a, the power of 4 with 5 as the base of log is 4, and the power of 12 with 25 as the base of log is represented by A.B

It is known that log5 (3) = a, log5 (4) = B,
The results are: log25 (12) = log5 & # 178; (3 × 4) = (1 / 2) log5 (3 × 4) = (1 / 2) [log5 (3) + log5 (4)] = (1 / 2) (a + b)

It is known that the quadratic power of (- 8 to the power of a) △ 4 to the power of 5-2B = 2 to the power of 1-B, the quadratic power of 9 to the power of a to the power of a to the power of B = 3 to the power of a to the power of 27
(1) Find the value of a to the second power + B to the second power
(2) Calculate [AB / (a + B-1) (a + b)] + [(AB / (a + b) (a + B + 1)] + [(AB / (a + B + 1) (a + B + 2)] +... + [(AB / (a + B + 99) (a + B + 100)]

1) (-8^a)^2÷4^(5-2b)=[2^(1-b)]^2
2^(6a)=2^(2-2b)*2^(10-4b)
2^(6a)=2^(12-6b)
6a=12-6b
a+b=2 ①
(9^a)^b=(3^b)^a÷27
3^(2ab)=3^(ab-3)
2ab=ab-3
ab=-3 ②
a^2+b^2=(a+b)^2-2ab=10
2) [ab/(a+b-1)(a+b)] +[ab/(a+b)(a+b+1)]+[ab/(a+b+1)(a+b+2)]+...+[(ab/(a+b+99)(a+b+100)]
=ab[1/(a+b-1)(a+b)] +[1/(a+b)(a+b+1)]+[1/(a+b+1)(a+b+2)]+...+[(1/(a+b+99)(a+b+100)]
=ab[1/(a+b-1)-1/(a+b)+1/(a+b)-1/(a+b+1)+1/(a+b+1)-1/(a+b+2)+...+1/(a+b+99)-1/(a+b+100)]
=ab[1/(a+b-1)-1/(a+b+100)]
=101ab/[(a+b-1)(a+b+100)]
=101*(-3)/[(2-1)*(2+100)]
=-101/34

3 and log3 4

0.3 ＾ 2, 0.3 ＾ 1, so 0.3 ＾ 2 ＾ 1
In log34, 4 > 3, so log34 > 1
0.3＾2＜log3 4

The 2n + 1 power of 14 - | (- 2) 2 | - (- 3) 3 ÷ (- 1) (n is a natural number)
Is the 2n + 1 power of (- 1)

The 2n + 1 power of 14 - | (- 2) 2 | - (- 3) 3 ÷ (- 1)
=14-4-27÷1
=10-27
=-17

How to calculate 24 minus 24 times 13 / 18

=24×(1-13/18)
=24×5/18
=20/3
=6 and 2 / 3

It is known that the equation of circle C is x2 + y2-2x + 4y-4 = 0, and the inclination angle of line L is 45 degrees
The first question is to find the linear equation if it is tangent
The second question is to find the maximum value of the triangle formed by the original center and the linear equation if two intersection points intersect

A:
If the inclination angle of the straight line is 45 °, the slope k = 1
Let the line be y = x + B
And the circular equation x ^ 2 + y ^ 2-2x + 4y-4 = 0
x^2+(x+b)^2-2x+4(x+b)-4=0
arrangement:
2x^2+(2b+2)x+b^2+4b-4=0
1）
A line is tangent to a circle, the intersection point is unique, and the equation has a solution
Discriminant = (2B + 2) ^ 2-4 * 2 (b ^ 2 + 4b-4) = 0
4b^2+8b+4-8b^2-32b+32=0
-4b^2-24b+36=0
b^2+6b-9=0
(b+3)^2=18
The solution is: B = - 3 + 3 √ 2 or B = - 3-3 √ 2
The linear equation is y = x-3 + 3 √ 2 or y = x-3-3 √ 2
2）
According to Weida's theorem:
x1+x2=-b-1
x1*x2=(b^2+4b-4)/2
Please confirm the center or origin?

Solve the equation X-5 of 12 = 5 of 730.4 △ (4-x) = 14.5 △ (x + 2) = 0.2 3x-4 times 6 = 4.8
Solve the equation X-5 of 12 = 5 of 730.4 △ (4-x) = 14.5 △ (x + 2) = 0.2 3x-4 times 6 = 4.8

What grades do you have on it
X-(12/5)=73/5
5X-12=73
X=17
The second one is whether there is a problem. The result is x = 139 / 35
Third
5/(x+2)=1/5
x+2=25
x=23
Fourth
(3x-4)6=4.8
3x-4=0,8
x=1.6

The proposition "there exists an X ∈ R, 2x ^ 2"_ If 3ax + 9 is less than 0, it is a false proposition

This proposition is a false proposition, which shows that 2x ^ 2_ No solution for 3ax + 9 less than 0
So we can get 2x ^ 2_ 3ax + 9 > = 0
We get 9A ^ 2-2 * 9 * 4

Simple 39 and 13 / 15 divided by 13

=(39+13/15)÷13
=39÷13+13/15÷13
=3+1/15
=3 and 1 / 15

The equation of a line tangent to the circle x2 + y2 = 4 and with a slope of - 1 is

k=-1
Then y = - x + B
Dai Ruyuan
2x²-2bx+b²-4=0
Tangency is a common point
So the equation has equal roots
So delta = 0
4b²-8b²+32=0
b=±2√2
So x + Y-2 √ 2 = 0 and X + y + 2 √ 2 = 0