Evaluate 2log &? 8323; 2-1og &? 8323; 32 / 9 + Log &? 8323; 8-5 &? 710; (2log5 3)

Evaluate 2log &? 8323; 2-1og &? 8323; 32 / 9 + Log &? 8323; 8-5 &? 710; (2log5 3)

1og₃32=5log₃2
log₃8=3log₃2
5ˆ（2log5 3）=8
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If log (8) (9) = A and log (3) (5) = B, LG2 is represented by a and B
The preceding brackets indicate what is the base

2/(2+3ab)
lg2=1/log(2)(10)=1/（log(2)(2)+log(2)(5)）
=1/{1+〖log(3)(5)/log(3)(2)〗}
=1/{1+〖log(3)(5)log(2)(3)〗}
=1/（1+b*(3/2)a）
=2/(2+3ab)

7x²+2x=0
Solve the equation
process

Solution equation: 7x & # 178; + 2x = 0
7x²+2x=0
Extracting the common factor X: X * (7x + 2) = 0
The above formula should be established,
Then: x = 0, or: 7x + 2 = 0
The results are as follows
X = 0, or x = - 2 / 7

If the fraction | a | / (A-A ^ 2) = 1 / (A-1), find the value range of A

The range of a is a < 0

Given the function f (x) = | x2-4x + 3 | (1) find the monotone interval of function f (x) and point out its increase and decrease; (2) if the equation f (x) - a = x about X has at least three unequal real roots, find the value range of real number a

（1）f（x）=|x2-4x+3|=x2−4x+3     (x≤1)−x2+4x−3    (1＜x＜3)x2−4x+3       When x ≤ 1, the function is a decreasing function; when 1 ≤ x ≤ 2, the function is an increasing function; when 2 ≤ x ≤ 3, the function is a decreasing function; when x ≥ 3, the function is an increasing function. From this, we can conclude that the monotone increasing intervals of the function are [1,2] and [3, + ∞], and the decreasing intervals are (- ∞, 1] and [2,3] (2) the equation f (x) - a = x, that is, f (x) = x + A, by y = x + A and y = - x2 + 4x-3, eliminating y, When a = - 34, the line y = x + A and the curve y = - x2 + 4x-3 are tangent to the point a (32, 34), and ∵ when the line y = x + a passes through the point B (1, 0), the two images also have three common points. At this time, a = - 1 ∵ when the line y = x + A is located between the points a and B (including the boundary), the two images have at least three different intersections. Therefore, a ∈ can be obtained by combining with the function image [-1，-34]．

The absolute value of K-4 of y = (K-2) x is an inverse proportional function. What is k equal to if (K-2) < 0

Absolute value of K - 4 = - 1
（k-2）＜0
k=-3

When x = | a | / A + | B | / B + | C | / C, try to find the value of - 92x + 2 to the power of 2003

∵ three rational numbers a, B, C whose product is negative and their sum is positive
There is a negative number and two positive numbers in it
∵x=|a|/a+|b|/b+|c|/c
∴x=1
The 2003 power of X - 92x + 2 = 1-92 + 2 = - 89

It is known that M + 1 + 2 of the square of - 5 / 1 x y + 6 of the square of - 4 x is a quaternion of degree 6
The degree of the third power of the monomial 4.5X and the fifth power of y-n is the same as that of this polynomial. We can find the square of M + the square of n
And explain why. Wait!

The degree of the third power of the monomial 4.5X and the fifth power of y-n is the same as that of this polynomial. We can find the square of M + the square of n
Write out the answer and the process, and explain it

6X minus 16 = 11x minus 34 (solving equation)
Solve it in equation format! Fast!

6x-16=11x-34
6x-11x=-34+16
-5x=-18
5x=18
x=18/5
x=3.6

If the inequality X & # 178; + 4x + 6-A ≥ 0 has a solution when - 3 ≤ x ≤ 1, find the value range of real number a

a∈(-∞,11]
x^2+4x+6 - a ≥ 0
Let f (x) = x ^ 2 + 4x + 6-a
∵△=b^2-4ac=4^2-4*1*(6-a)=16-24+4a=4a - 8
∴x= [-4±2√(a-2)]/2= -2±√(a-2)
The solution of the original inequality is as follows:
-2 - √(a-2) ≤ x ≤ -2+√(a-2)
∴-2 - √(a-2) ≥ - 3
√(a-2) ≤ 1
a-2 ≤ 1
a ≤ 3
-2+√(a-2) ≤ 1
√(a-2) ≤ 3
a-2 ≤ 9
a ≤ 11
∴a∈(-∞,11]