Solving the equation about X: A / A-X + a = 1

Solving the equation about X: A / A-X + a = 1

a/(a-x)+a=1
a+a(a-x)=a-x
a+a²-ax=a-x
(1-a)x=-a²
x=-a²/(1-a)

If the equation x = a + 1 and 2 (x-1) = 5a-6 have the same solution, then a = ()

Because the equation x = a + 1 about X has the same solution as 2 (x-1) = 5a-6
So substituting x = a + 1 into 2 (x-1) = 5a-6, we get
2 (a + 1-1) = 5a-6, that is, 2A = 5a-6
The solution is a = 2

Given the function f (x) = (A-1 / 2) x & # 178; + ln X. (a ∈ R) (1) when a = 1, find the maximum and minimum values of F (x) in the interval [1, e];
(2) Finding the extreme value of F (x)

Let x1, X2 ∈ [1, e], and satisfy x2 ＞ x1, and a = 1, then f (x2) - f (x1) = (1-1 / 2) [(x2) & # 178; - (x1) & # 178;] + (LN X2 - ln x1) = (1 / 2) (x2 + x1) (X2 - x1) + ln (x2 / x1) according to the property of logarithmic function y = LNX, the domain of definition of the original function f (x) is X

A number plus the product of this number and its reciprocal is 4 and 1 / 3. The reciprocal of this number is ()

I think it should be three tenths. The general approach is as follows:
First, the product of a number and its reciprocal is 1,
Let this number be x, then x + 1 = 4 and 1 / 3, then we can find that x is three tenths

4Y ^ 2 + 4y-1 = 0 solved by formula method

4y^2+4y-1=0
4y^2+4y=1
4y^2+4y+1=2
(2y+1)^2=2
2y+1=±√2
2y=-1±√2
y=(-1±√2)/2
Y = (√ 2-1) / 2 or y = (- √ 2-1) / 2

The number of zeros of degree x of function f (x) = x & # 178; - 2 is? A 3, B 2, C 1, d 0

What is the number of zeros of F (x) = x & # 178; - 2 ^ x?
Because it is zero, that is, f (x) = 0, that is, X & # 178; - 2 ^ x = 0,
I.e. X & # 178; - 2 ^ x = 0, i.e. X & # 178; = 2 ^ x,
That is to find the number of intersections of the functions y = x and y = 2 ^ X,
By drawing and combining number with shape, the number of intersections of two function images is obviously only 2

The parabola y = MX ^ 2 - (2m + 2) X-1 + m is known,
(1) When m is the value, there is only one intersection point between the parabola and the x-axis
(2) When m is the value, the parabola intersects with X axis
If you want to use a picture, you'd better draw a picture, or a sketch,

y=mx²-(2m+2)x-1+m
Since it has been said that it is a parabola, then m ≠ 0
(1)
There is only one intersection point between the parabola and the x-axis
Then Δ = (2m + 2) &# 178; - 4m (- 1 + m) = 12m + 4 = 0
So m = - 1 / 3
(2)
The parabola has an intersection with the x-axis
Then Δ = 12m + 4 ≥ 0
So m ≥ - 1 / 3
And m ≠ 0
So {m | m ≥ - 1 / 3 and m ≠ 0}
If you don't understand, please ask, I wish you a happy study!

The vertex of the parabola y = x ^ 2-2x + m is on the x-axis
Marching

The vertex on the x-axis represents Δ = 0 and 4-4m = 0 to get m = 1. At this time, we can use the formula to get the vertex coordinates as follows: when y = 0 is x = 1 and (1,0)
So the axis of symmetry is x = 1
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One by half plus two by three plus three by four until forty-nine by fifty

=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+…… +(1/49-1/50)
Middle positive and negative offset
=1-1/50
=49/50

A. the number whose absolute value is equal to 3 is - 3 B. integers whose absolute value is less than 1 and 1 / 3 are 1 and - 1 C. the number with the smallest absolute value is rational
The absolute value of D.3 is 3

D. The absolute value of 3 is 3