Solving fractional equation: 1 / (x + 2) + 1 / (x + 7) = 1 / (x + 3) + 1 / (x + 6)

Solving fractional equation: 1 / (x + 2) + 1 / (x + 7) = 1 / (x + 3) + 1 / (x + 6)

General division
(x+7+x+2)/(x+2)(x+7)=(x+6+x+3)/(x+3)(x+6)
(2x+9)/(x^2-9x+14)-(2x+9)/(x^2+9x+18)=0
(2x+9)[1/(x^2-9x+14)-1/(x^2+9x+18)]=0
Because x ^ 2-9x + 14 is not equal to x ^ 2 + 9x + 18
So 1 / (x ^ 2-9x + 14) - 1 / (x ^ 2 + 9x + 18) is not equal to 0
So 2x + 9 = 0
x=-9/2
The fractional equation needs to be tested
After inspection
X = - 9 / 2 is the solution of the equation
(x+7+x+2)/(x+2)(x+7)=(x+6+x+3)/(x+3)(x+6)
(2x+9)/(x^2-9x+14)-(2x+9)/(x^2+9x+18)=0
(2x+9)[1/(x^2-9x+14)-1/(x^2+9x+18)]=0
Because x ^ 2-9x + 14 is not equal to x ^ 2 + 9x + 18
So 1 / (x ^ 2-9x + 14) - 1 / (x ^ 2 + 9x + 18) is not equal to 0
So 2x + 9 = 0
x=-9/2
1/(x+2) + 1/(x+7) = 1/(x+3) + 1/(x+6)
1/(x+2) - 1/(x+3) = 1/(x+6)- 1/(x+7)
General score:
1/[(x+2)(x+3)]=1/[(x+6)(x+7)]
(x+2)(x+3)=(x+6)(x+7)
x^2+5x+6=x^2+13x+42
8x=-36
x=-9/2
Solve the fractional equation: 1 / (x-3) + 1 / (X-7) = 1 / (x-4) + 1 / (X-6)
1/(x-3)+1/(x-7)=1/(x-4)+1/(x-6).
1/(x-3)-1/(x-4)=1/(x-6)-1/(x-7)
(x-4-x+3)/(x-3)(x-4)=(x-7-x+6)/(x-6)(x-7)
-1/(x-3)(x-4)=-1/(x-6)(x-7)
(x-6)(x-7)=(x-3)(x-4)
x²-13x+42=x²-7x+12
-6x=-30
X=5
Test: x = 5 is the root of the equation
If the fractional equation x + 2 / x-x-2 / x + 2 = x & # 178; - 4 / k about X has no solution, find the value of K
x+2/x-x-2/x+2=x²-4/k
x²=2+4/k
Because the equation has no solution, so 2 + 4 / K
If there is no solution to the fractional equation 1 / (X-2) + K / (x + 2) = 3 / (x ^ 2-4) of X, find the value of K
Solving equation (x + 3) / (x + 4) = (x + 6) / (x + 5)
Two sided multiplication x ^ 2-4 (x + 2) + K (X-2) = 3x + KX + 2-2k = 3 (K + 1) x = 1 + 2kx = (2k + 1) / (K + 1) has no solution. In one case, if (K + 1) x = 1 + 2K, K + 1 = 0, then 0 * x = - 1 does not hold. In the second case, if there is an increasing root, that is, the denominator is 0, so x = 2, x = - 2x = (2k + 1) / (K + 1) = 22K + 1 = 2K + 2, x = (2k + 1) / (K + 1) = -
No matter x is any real number, the value of the algebraic formula 2x - 4x + 5 is always positive
2x^2-4x+5
=2x^2-4x+2+3
=2(x^2-2x+1)+3
=2(x-1)^2+3
No matter what value x takes, there is (x-1) ^ 2 ≥ 0
So no matter x is any real number, 2 (x-1) ^ 2 + 3 is greater than 0
No matter x is any real number, the value of the algebraic formula 2x-4x + 5 is always positive
Original formula = 2 (x ^ 2-2x + 2.5)
=2(x^2-2x+1+1.5)
=2(x-1)^2+3
∵(x-1)^2≥0，
∴2(x-1)^2≥0，
∴2(x-1)^2+3≥3
The value of the original algebra is always positive
If X - (5 + 2Y) = 15, then 2x-4y =?
x-(5+2y)=15
x-2y-5=15
x-2y=20
2x-4y=2(x-2y)=2×20=40
It is known that if x-2y = 20, then 2x-4y = 2 * (x-2y) = 40
Both sides of the equation multiply by 2, that is, 2x-4y = 15 + 10 = 25
Forty
x-(5+2y)=15
x-5-2y=15
x-2y=20
2(x-2y)=2x-4y=2×20=40
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
Solve the following equation: 3Y + 5 / 3y-6 = 1 / 2 + 5y-4 / 2y-4; 1 / x + 4-1 / x + 7 = 1 / x + 3-1 / x + 6
(3y+5)/(3y-6)=1/2+(5y-4)/(2y-4)
2 (3Y + 5) = 3 (Y-2) + 3 (5y-4)
6y+10=3y-6+15y-12
6y-3y-15y=-6-12-10
-12y=-28
y=7/3
It is the root of the original equation
1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)
1/(x+6)-1/(x+7)=1/(x+3)-1/(x+4)
[(x+7)-(x+6)]/(x+6)(x+7)=[(x+4)-(x+3)]/(x+3)(x+4)
1/(x+6)(x+7)=1/(x+3)(x+4)
(x+6)(x+7)=(x+3)(x+4)
x²+13x+42=x²+7x+12
13x-7x=12-42
6x=-30
x=-5
It is the root of the original equation
Given that the point P is the moving point on the straight line x + y + 6 = 0, PA and Pb are the two tangents of the circle x2 + y2-2x-2y + 1 = 0, a and B are the tangent points, and C is the center of the circle, then when the area of the quadrilateral PACB is the smallest, the coordinates of point P are______ .
According to the product of tangent length and radius is the area of triangle, when the area of quadrilateral PACB is the smallest, the tangent length is the smallest, the distance of PC is the smallest, and the straight line equation passing through the center of circle and perpendicular to the straight line is Y-1 = X-1, that is y = x, so y = XX + y + 6 = 0, the solution is x = y = - 3, so the coordinates of point P are: (- 3, - 3), so the answer is: (- 3, - 3)
It is proved by collocation that the value of the algebraic formula 2x-x2-5 is always less than zero
It is proved by collocation that the value of the algebraic formula 2x-x2-5 is always less than zero
2x-x²-5
=-x²+2x-5
=-(x²-2x+1)-4
=-(x-1)²-4
∵ nonnegative square,
∴ -(x-1)²≤0
∴ y=-(x-1)²-4