The development of exponential functions, exponents and exponential functions

First, in order to simplify the onerous four-rule arithmetic, the logarithm was invented, then the logarithmic function was invented, and then the inverse function was invented.

Here 2 (-x)=1 2 x, and then pass the divide, and the numerator denominator is about 2 x to get this equation.

Because 2 -x = 1 (2 x), this is the formula, and then it is the simplest elementary school knowledge.

The numerator and denominator are multiplied by 2 x at the same time, 1-2 (-x)] 1+2 (-x)]=2 x-2 (-x)*2 x] [2 x+2 (-x)*2 x].

2 x-1) (2 x+1), note 2 (-x)*2 x=2 0=1

The above numerator and denominator are multiplied by 2 x at the same time, and the following equation can be simplified.

f(x)=(3 2x)-[k+1)3 x]+2(3 x) 2 - k+1)*(3 x) +2Let 3 x = t

f(t) =t 2 - k+1)t + 2x belongs to r then the finch cavity t > 0

The problem translates to : t>0, f(t) is always positive f(t) =t 2 - 2*[(k+1) 2]x + k+1) 2] 2 - k+1) 2] 2] 2 + 2

t - k+1) 2] 2 + 8 - k+1) 2] 4f(t) is a parabola with t = k+1) 2 as the axis of symmetry and [8 - k+1) 2] 4 as the minimum.

In order to ensure that f(t) is always positive when t>0, it is required that 1) the minimum value is greater than 0

or 2) the minimum value is less than 0, but the axis of symmetry is in the range of t < 0, and f(0) 0

For case 1), then.

8 - k+1)^2 > 0

1 - 2 2 < k < 1 + 2 2 for case 2) is required.

k+1)/2 < 0

f(0) =2 > 0

Solve k < 1

Take the above two solutions and set leather shirts, then the years are wild.

k < 1 + 2√2

1, the numerator and denominator are multiplied (1-2 (-1 32)) and the purification of the numerator is 1 2, and the denominator is (1-2 (-1 32)).

2，（x^(1/2)+x^(-1/2)）^3=x^(3/2)+3*x^(1/2)+3*x^(-1/2)+x^(-3/2)=27

Therefore, the positive branch x (3 2) + x (-3 2) = 27-3 * 3 = 183I don't quite understand this.

Analysis: This question uses the cube sum formula which book is worse.

A 3 + B 3 = A+B) (A 2-Ab+B 2) Solution: (Lg2) 3+(Lg5) 3+3*LG2*LG5 Anterior Plum Peel Face Pose is dismantled with the sum of cubes.

lg2+lg5)[(lg2)^2-lg2*lg5+(lg5)^2]+3*lg2*lg5

lg10[(lg2)^2-lg2*lg5+(lg5)^2]+3*lg2*lg5

1*[(lg2)^2-lg2*lg5+(lg5)^2]+3*lg2*lg5

lg2)^2+2*lg2*lg5+(lg5)^2(lg2+lg5)^2

lg10)^2

So a+b=1

Hence a 3 + b 3 + 3 ab

a+b)[a^2-a*b+b^2]+3ab1*(a^2-a*b+b^2)+3ab

a^2+2a*b+b^2

a+b)^2