If the fractional equation x / X-1 + K / X-1 = 2 has an increasing root x = 1, find the value of K

If the fractional equation x / X-1 + K / X-1 = 2 has an increasing root x = 1, find the value of K

Take (x + k) = 2 (x-1) into x = 1
k=-1
x+k=2x-2
x=k+2=1
k=-1
For non-zero natural numbers m and N, the meaning of "*" is: m * n = a times m + N divided by 2 times m times n (a is a certain integer). If 3 * 6 = 4 * 5
So 4 * 6 = ()
3*6=3A+6÷2
4*5=4A+5÷2
Because the two are equal
So 3A + 6 △ 2 = 4A + 5 △ 2
The solution is a = 0.5
So 4 * 6 = 4 × 0.5 + 6 △ 2 = 5
Change the order of integration ∫. (- A, a) DX ∫ (0, (radical a ^ 2-x ^ 2)) f (x, y) dy
∫(-a,a)dx∫(0,√(a^2-x^2)f(x,y)dy=∫(0,a)dy∫(-√(a^2-y^2),√(a^2-y^2)f(x,y)dx.
Solve the equation 1 - 2x-56 = 3-x4
1- 2x-56 = 3-x4
-2x+4x=3-1+56
2x=58
x=58/2
1-2x-53=-4x
1+2x=53
2x=54
x=27
29
1- 2x-56 = 3-x4
Transfer: 4x-2x + 1-56-3 = 0
Merge similar items: 2x-58 = 0
2X=58
X=58/2
X=29
1-2x-56=3-x4
4x-2x=3-1+56
2x=58
x=29
Let f (x) be a function defined on the real number set R, satisfying f (0) = 1, and for any real number a and B, f (a) - f (a-b) = B (2a-b + 1), then the analytic expression of F (x) can be ()
A. f（x）=x2+x+1B. f（x）=x2+2x+1C. f（x）=x2-x+1D. f（x）=x2-2x+1
(paper) a [analysis]: let a = b = x, then f (x) - f (0) = x (2x-x + 1) = x2 + X. then f (0) = 1, ｜ f (x) = x2 + X + 1
Calculate the multiple integral ∫ (lower 0, upper 1) DX ∫ (lower 0, upper √ x) e ^ (- y ^ 2 / 2) dy
Exchange the order of integration, and then use partial integration, as follows:
Solving the equation: 1 / 4 x + 2 / 7 x = 56
(1/4) X+(2/7)X=56
(15/28)*X=56
X = (1568/15)
We know that the definition of function f (x) = x ^ ((1-A) / 3) is a non-zero real number, and it is an increasing function on (- infinity, 0) and a decreasing function on (0, positive infinity),
The domain of definition is all real numbers. When x = 0, y is always equal to 0, but we can't say Y / x = k, because y / X is an indefinite number, so we can only use y = KX
Calculate the integral ∫ (0,2) DX ∫ (x, 2) e ^ (- Y & # 178;) dy
Draw the integral area D and change the integral order
∫(0～1)dx∫(x～1)e^(-y^2)dy
＝∫(0～1)dy∫(0～y)e^(-y^2)dx
The original function of = ∫ (0 ～ 1) Ye ^ (- y ^ 2) dy integrand is - 1 / 2E ^ (- y ^ 2)
＝1/2×(1－1/e)
＝(e－1)/(2e)
If y equals 5x minus 1 under the root sign and then 1 minus 5x under the root sign, what is the sum of 5x plus y
y=√(5x-1)+√(1-5x)
From the condition that quadratic radical is meaningful, 5x-1 ≥ 0, and 1-5x ≥ 0
So, 1-5x = 0
Then, x = 1 / 5
In this case, y = 0
So, 5x + y = 5 * (1 / 5) + 0 = 1
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