Solving logarithmic equation lg[(1-x)/(1+x)]+1/(x+2)=0

Solving logarithmic equation lg[(1-x)/(1+x)]+1/(x+2)=0

The way to solve the exponential equation is to remove the exponential expression and change it into an algebraic equation
In this way, solving the exponential equation is the problem of transforming the exponential form
There are three types of questions, as follows
1. A ^ [f (x)] = type B
Change to logarithm
Then a ^ [f (x)] = B;
2. A ^ [f (x)] = a ^ [g (x)] type: F (x) = g (x);
3. Univariate quadratic form: a [a ^ f (x)] & sup2; + Ba ^ f (x) + C = 0
Let a ^ f (x) = t (where t > 0)
In some extra-curricular books, there are questions like a ^ x = x + 1. This kind of question uses the image method. In the same coordinate system, draw the image of exponential function and first-order function respectively, and see that the number of intersections is the number of equation roots. Generally, the exact value can not be obtained
You'd better try drawing

The area of a square is 10 square centimeters. How many square centimeters is the area of the painted part?

14 × 10 × 34, = 23.55 (square centimeter); a: the area of shadow is 23.55 square centimeter

One day, the circus was going to hold an animal sports meeting, and coke broke the animals. At the beginning of the competition, the elephant referee announced that the first dog and monkey would take part in the 100 meter preliminary race. Unexpectedly, when the dog ran to the end, the monkey ran to 90 meters, and his mouth pouted out in anger! In the final, the clever monkey suddenly put forward: "the dog is born to run fast, If we stand on the same starting line, it's not fair to race. I suggest that its starting line move back 10 meters. "The dog holds the monkey's hand and agrees. The monkey thinks happily, so that the dog and I will reach the finish line at the same time
Will the monkey get what he wants?

No, you can set the dog's speed to V1 and the monkey's speed to v2
So 90 / V2 = 100 / v1
We get V1 = (10v2) / 9
The second game begins,
Let's set the time for the dog as T1 and the time for the monkey as T2
So T1 = 110 / V1 = 110 / (10v2 / 9) = 90 / V2
T2=100/v2
T2 was significantly greater than T1, that is, the monkey used a long time, the dog first to the end

If the polynomial 2x ^ n + (m-1) X-1 is a cubic binomial about X, find the value of m ^ 2-N ^ 2

2X ^ n + (m-1) X-1 is a cubic binomial of X
∴n=3
m-1=0,m=1
∴m²-n²
=(m+n)(m-n)
=(3+1)(3-1)
=4×2
=8

The number of roots of the equation x · LG (x + 1) = 1 is______ One

Equation x · LG (x + 1) = 1, that is, LG (x + 1) = 1 x, so the number of roots of equation x · LG (x + 1) = 1 is equal to the number of intersections of function y = LG (x + 1) and function y = 1 X image, as shown in the figure: since function y = LG (x + 1) and function y = 1 x image have 2 intersections, the number of roots of equation x · LG (x + 1) = 1 is 2, so the answer is 2

Fifth grade disjunctive calculation problem ~ urgent ~ urgent ~ urgent ~ (want answers)
More than 20, thank you (with answers)

2/1*2=1 3/1*3=1 3/2*3=2 3/1*6=2 4/3*8=6 5/3*20=12 7/3*14=6 8/7*40=35 4/3*16=12 9/5*27=15 2/1*30=15 12/7*24=14 30/1*30=1 51/9*102=18 19/9*76=36 4/9*8=18 5/8*90=144 99/98*99=98 3/14*6=28 7/1*28=4 10/1*9...

Y = LNX, P is a point on the function, O is the origin of the coordinate, and the maximum slope of OP is calculated. The usual way is to derive k = LNX / X. I will, but I have a question. From the image, the tangent point of the function and the straight line is the maximum value of K, and the point should be the maximum value of G = LNX KX. How to solve this problem

Your direction is feasible, the process of transformation is wrong
The problem of G (x) = LNX KX, x > 0 is transformed into the equation g (x) = 0 and the maximum value of K is obtained
g'(x)=1/x-k=(1-kx)/x.
When k ≤ 0, G '(x) > 0, G (x) increases monotonically, and G (x) = 0 has a solution
k> Let g '(x) = 0, then x = 1 / K
When 0

Given that the lengths of the three sides of the angle ABC are a, B, C, and a ^ 2 + B ^ 2 + C ^ 2 = AB + BC + Ca, then the angle ABC is ()
A arbitrary triangle B equilateral triangle C isosceles triangle D right triangle

The answer is ba & sup2; + B & sup2; + C & sup2; = AB + BC + Ca ﹣ 2A & sup2; + 2B & sup2; + 2C & sup2; = 2Ab + 2BC + 2ca ﹣ 2A & sup2; + 2B & sup2; + 2C & sup2; - 2ab-2bc-2ca = 0 ﹣ (a-b) & sup2; + (B-C) & sup2; + (A-C) & sup2; = 0 ﹣ (a-b) & sup2; ≥ 0 (B-C) & sup2; ≥ 0 (A-C)

What is the simplest calculation formula of CPK?

　　0.1639 0.2948 0.2881
　　0.1603 0.2902 0.2923
　　0.1707 0.2933 0.2926
　　0.1646 0.2911 0.2859
　　0.1748 0.3023 0.2921
　　0.1575 0.2994 0.2969
　　0.1531 0.2985 0.2926
　　0.1661 0.2996 0.2868
　　0.166 0.2968 0.2931
　　0.1666 0.2935 0.2884
　　0.1695 0.3033 0.2968
　　0.1622 0.297 0.2919
　　0.1732 0.2941 0.294
　　0.1687 0.2999 0.2986
　　0.1676 0.2918 0.2911
　　0.1671 0.2893 0.2897
　　0.1601 0.2977 0.2926
　　0.1605 0.3047 0.2911
　　0.1671 0.2947 0.2928
　　0.1674 0.296 0.287
　　0.1704 0.2983 0.2939
　　0.1687 0.2967 0.2856
　　0.1588 0.2922 0.2942
　　0.1634 0.2951 0.2896
　　0.1694 0.2942 0.2894
　　0.1681 0.2934 0.2894
　　0.1661 0.2918 0.2844
　　0.1617 0.2957 0.2875
　　0.1669 0.292 0.289
　　0.168 0.2879 0.2915
Average 0.165616667 0.29551 0.290963333 CPK = min (usl-x or x-lsl)
　　sigma 0.004753807 0.004080259 0.003452584 3σ
　　3sigma 0.01426142 0.012240778 0.010357751
　　σ = R
D2 sampling constant, look up the table
　　USL 0.168 0.303 0.303
　　LSL 0.162 0.27 0.27
　　B31 - B37 0.003616667 0.02551 0.020963333
　　B36- B31 0.002383333 0.00749 0.012036667
　　CPK 0.167117532 0.611889193 1.162092716

9-3x=2x-2

9-3x=2x-2
5x=11
x=11/5