If the fractional equation 2 / X-2 + KX / x ^ 2-4 = 3 / x + 2 has no solution, then the value of K is Please write down the detailed process, good person o (∩)_ Thank you It's clear. 2/(x-2)+kx/(x^2-4)=3/(x+2)

If the fractional equation 2 / X-2 + KX / x ^ 2-4 = 3 / x + 2 has no solution, then the value of K is Please write down the detailed process, good person o (∩)_ Thank you It's clear. 2/(x-2)+kx/(x^2-4)=3/(x+2)

Multiply both sides by (x ^ 2) - 4
We obtain 2 (x + 2) + KX = 3 (X-2) ①
According to the property of fractional equation, if it has a solution, then x ≠ 2, - 2
If we want him to have no solution, then x = 2, - 2
Substituting into equation: i.e
x=2：8+2k=0,k=-4
x=-2：-2k=-12,k=6
If the fractional equation x + 1 / (X & # 178; - x) - 1 / 3x = K / 3x-3 about X has no solution, find the value of K
(x+1)/(x^2-x)=1/3x+(x+k)/(3x-3)
(x+1)/x(x-1)=1/3x+(x+k)/3(x-1)
If there is no solution, the denominator is 0
3x(x-1)
X = 0 or x = 1
3(x+1)=(x-1)+(x+k)x
3x+3=x-1+x^2+kx
x^2+(k-2)x-4=0
So 0,1 are the two roots of the equation
So 0 + 1 = - (K-2)
=>k=1
When the value of K is what, there is no solution to the fractional equation x + 1 / K plus X-1 / 1 = x minus 1 / 1?
De denominator (x + 1) (x-1)
kx-k+x+1=1
(k+1)x=k
X = - 1, no solution
k≠-1
Then x = K / (K + 1) is an increasing root
The denominator is 0
x=1,x=-1
K / (K + 1) = 1, k = K + 1, not true
k/(k+1)=-1,k=-k-1,k=-1/2
So k = - 1, k = - 1 / 2
It should be discussed according to the situation
1, when k = 0, there is no solution
2, when k ≠ 0
k(x+1)+(x-1)=x+1
k(x+1)=0
x=-1
Substituting x = - 1 into the original fraction, increasing the root, there is no solution
Given x + 2Y = 3, x ^ 2-4y ^ 2 = - 15, find the value of x-2y. Given x + 2Y = 3, x ^ 2-4y ^ 2 = - 15. (1) find the value of x-2y. (2) find the value of X and y
X-2Y=0
X ^ 2-4y ^ 2 decomposes (x + 2Y) (x-2y), because x + 2Y = 3, so x-2y = - 5, according to x-2y = - 5, x + 2Y = 3, so x = - 1, y = 2
5) In the quadratic function y = ax & sup2; + BX + C, when x = 1, the minimum value of Y is - 2, and the image passes through (- 3,4), find a, B
5) Given the quadratic function y = ax & sup2; + BX + C, when x = 1, the minimum value of Y is - 2, and the image passes through (- 3,4), find the value of a, B, C
(6) It is known that a (x1,.), B (x2.0) are on the parabola y = x & sup2; + MX + NX, and x1.x2 is the two roots of the equation x & sup2; - 4x + 3 = 0. The relation of this parabola is obtained
1>
solution
Because when x = 1, y has a minimum of - 2
So B = 1 / 2A
-2 = a+ b +c ( 2)
4 = 9a -3b +c (3)
a= 3/8 b= -3/4 c = -13/8
2>
A (x1,.), B (x2.0) according to a (x1,) y = x & sup2; + MX + nx
Is there a problem? Is the original copy wrong
What is (5-3y) + 2Y = 15 - (7-5y) equal to?
(5-3y)+2y=15-(7-5y)
5-3y+2y=15-7+5y
5-y=8+5y
6y=-3
y=-1/2
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(5-3y)+2y=15-(7-5y)
5+(-3y+2y)=(15-7)-5y
5-y = 8-5y
5y-y=8-5
4y=3
y=3/4
5-3y+2y=15-7+5y
-3y+2y-5y=15-7-5
-6y=3
y=-3/6
What a long time ago
5-3y + 2Y = 15-7 + 5Y
5 -y=8+5y
-y-5y=8-5
-6y=3
y=-1\2
It is known that P is the moving point on the straight line 3x + 4Y + 8 = 0, PA and Pb are the two tangent lines of the circle x2 + y2-2x-2y + 1 = 0, a and B are the tangent points, C is the center of the circle, then the minimum area of PACB of quadrilateral is______ .
∵ the equation of circle is: x2 + y2-2x-2y + 1 = 0 ∵ center C (1,1), radius R is: 1. According to the meaning, if the area of quadrilateral is the smallest, when the distance between the center of circle and point P is the smallest, and the distance between the center of circle and straight line is the distance between the center of circle and straight line, the tangent length PA, Pb is d = 3 ∵ PA | = | Pb | = D2 − R2 = 22 ∵ spacb = 2 × 12 | PA | r = 22, so the answer is: 22
Use the collocation method to find the maximum value of - 2x & # 178; + 8x - 10
-2x²+8x-10
＝-2(x²-4x+4)-2
＝-2(x-2)²-2.
When x = 2,
The maximum value is: - 2
If x and y satisfy x2 + y2-2x + 4Y = 0, then the maximum value of x-2y is ()
A. 0B. 5C. -10D. 10
First, according to x, y satisfying x2 + y2-2x + 4Y = 0, we can draw a graph on the circle with (1, - 2) as the center and 5 as the radius. Let z = x-2y, then When the line z = x-2y passes through the intersection a (2, - 4) of the line OC and the circle, the intercept − Z2 on the y-axis is the minimum and Z is the maximum. Substituting the point a (2, - 4) into Z = x-2y, the maximum value of Z is: 10. So the maximum value of x-2y is 10
Given the quadratic function y = AX2 + BX + C, when x = - 1, there is a minimum value of - 4, and the length of the line segment of the image on the x-axis is 4, the analytic expression of the function is obtained
∵ the symmetry axis of the parabola is x = - 1, the length of the line segment on the x-axis is 4, the coordinates of the intersection of the parabola and the x-axis are (- 3,0), (1,0), let the analytical formula of the parabola be y = a (x + 3) (x-1), and substitute the vertex coordinates (- 1, - 4) to get a (- 1 + 3) (- 1-1) = - 4, and the solution is a = 1, ∵ the analytical formula of the parabola is y = (x + 3) (x-1), that is y = x2 + 2x-3