It is known that the equation x ^ A + 2 + 3Y ^ 1-2b = 15 about X and Y is a quadratic equation of two variables. Find the value of a and B,

It is known that the equation x ^ A + 2 + 3Y ^ 1-2b = 15 about X and Y is a quadratic equation of two variables. Find the value of a and B,

The equation x ^ A + 2 + 3Y ^ 1-2b = 15 of X, y is a quadratic equation of two variables,
therefore
a+2=1
1-2b=1
therefore
a=1-2=-1
b=0

It is known that x ∧ a-2b and - 3Y ∧ 3a-2b-10 are quadratic equations with respect to X and y, and the value of A-B is obtained

Because the title is determined to be a quadratic equation of two variables, the powers of X and y are all 1. The problem is transformed into solving the equations of constant a and B, that is:
A-2b = 1, 3a-2b-10 = 1, the solution is a = 5, B = 2, so A-B = 3

A = Log1 / 2 3 B = 2 ^ 0.1 C = 3 ^ - 0.1 compare size

c∈(0,1),b>1,a

Compare the size of (Log1 / 4) 8 / 7, (Log1 / 5) 6 / 5

Log (1 / 4) 8 / 7 = LG (8 / 7) / log (1 / 4) = - LG (8 / 7) / LG4 = LG (7 / 8) / LG4. Similarly, log (1 / 5) 6 / 5 = LG (5 / 6) / lg57 / 8 > 5 / 6 leads to LG (7 / 8) > LG (5 / 6) lg5 > LG4 > 01 / LG4 > 1 / lg5lg (7 / 8) / LG4 > LG (5 / 6) / lg5 leads to log (1 / 4) 8 / 7 > log (1 / 5) 6 /

Evaluation: - log2log2 √ √ 2, log4 (8) - Log1 / 9 (3) - log √ 2 (4); (log4 (3)) * (log9 (25)) * (log5 (8))

-log2log2√√√2
=-1 / 8 power of log2
=-log2 (1/8)
=log2 8
=3
log4 (8)-log1/9 (3)-log√2 (4)
=(3 / 2) power of log4 - 1 ÷ [log3 (1 / 9)] - fourth power of log √ 2 √ 2
=3 / 2 - 1 △ [log3 (the - 2nd power of 3)] - 4
=3/2 + 1/2 -4
=2
(log4(3))*（log9(25))*(log5(8))
=(lg3/lg4)*(lg25/lg9)*(lg8/lg5)
=lg3/(2lg2)*(lg5/lg3)*(3lg2/lg5)
=3/2

The solution of log4 ^ (13-3x) * log (x-1) ^ 2 = 1

log4^(13-3x)*log(x-1)^2=1
log4^(13-3x)=1/log(x-1)^2
1/2*log2^(13-3x)=log2^(x-1)
So √ (13-3x) = X-1
square
13-3x=x^2-2x+1
x^2+x-12=0
x=-4,x=3
True number 13-3x > 0, X-1 > 0
So x = 3

Let f (x) = x (x-1) (X-2) (x-3), then: what is the number of roots of the equation f '(x) = 0 in [0,3]?

Because the result of F (x) = 0 is 0 1 2 3
Moreover, f (x) is continuous in the whole interval, which can be obtained from Rolle's theorem
In (0,1) (1,2) (2,3), each point satisfies f '(x) = 0, so it is 3

F (x) = log2 (2-x) + log2 (2 + x), the domain of definition is (- 2,2), the number of real roots of the equation f (x) = | x | is pointed out, and the reason is given

f(x)=log2 (2-x)+log2 (2+x)
=log2(-x^2+4)
Defined by (- 2,2)
Know 0

The function f (x) = x2-ax + 3 has f (x) > = a on [- 2,2]. How to find the value range of a?
rt

When the axis of symmetry A / 2F (x) = x2-ax + 3 is [- 2,2], 2 > = A / 2 > = - 24 > = a > = - 4f (x) = [x2-ax + 3] min = f (A / 2) = 3-A ^ 2 / 2 > = AA > = - 1 + 2 radical 2 or less than or equal to - 1-2 radical 2A belongs to [- 4, - 1-2 radical 2] and [- 1 + 2 radical 2,4] f (x) = x2-ax + 3 when the axis of symmetry is [2, positive infinity]

Given the function f (x) = x 2 + ax + 1, f (x) always holds f (x) ≥ - 3 on X ∈ [- 3,1), the value range of real number a is obtained

Because f (x) = x2 + ax + 1 = (x + A2) 2 + 1 − A24 (I) when − A2 ＜ 3, i.e. a ＞ 6, it is easy to know that it is an increasing function on X ∈ [- 3, 1), then f (x) min = f (− 3) = 10 − 3a ≥ − 3 {a ≤ 133, then a has no solution; (II) when − 3 ≤ − A2 ＜ 1, i.e. - 2 ＜ a ≤ 6, then f (x) min = f (− A2) = 1 − A24