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How to solve the equation (LNX) ^ 3-1 + x = 0?
Let f (x) = (LNX) & # 179; + X-1
Then the function f (x) increases on (0, + ∞) and:
If f (1) = 0, then:
The root of equation (LNX) &# 179; - 1 + x = 0 is x = 1
Equation LNX = x
There is no intersection between y = LNX and y = X
So there is no solution
Solving the equation: LNX + (x + 1) × 1 / X - 1 = 0 requires detailed process, thank you
Given that the root of X + e ^ x = 5 is a and the root of equation x + LNX = 5 is B, then a + B
E ^ x = 5-xlnx = 5-xy = e ^ X and y = LNX are inverse functions of each other, so the image is symmetric with respect to quadrant 1 and quadrant 3, that is, the line y = x is symmetric, the line y = 5-x is perpendicular to y = x, the intersection abscissa of Y = 5-x with y = e ^ X and y = LNX is the root of the above two equations, a and B. the two points of a and B are symmetric with respect to y = x, and the midpoint of a and B is y = x and y = 5
Given that the intersection of two lines L1: y = KX + 1 + 2k and y = - 1 / 2x + 2 is above the line y = x, then the range of K is larger
L1: Y-1 = K (x + 2) crossing point (- 2,1)
L2: the slope is - 1 / 2, which can be drawn
The intersection of L2 and y = x is (4 / 3,4 / 3)
So if we take the intersection point into Li, we can get a k = 1 / 10, so a range of K is (1 / 10, + ∞)
Because L1 and L2 intersect, K is not equal to - 1 / 2, and because L1 and L2 intersect above y = x, another range of K is
（-∞,-1/2).
If the intersection of two lines y = KX + 2K + 1 and X + 2y-4 = 0 is in the fourth quadrant, then the value range of K is ()
A. （-6，2）B. (−16，0)C. (−12，−16)D. (12，+∞)
The simultaneous equation y = KX + 2K + 1x + 2Y − 4 = 0 can be solved to x = 2 − 4k2k + 1y = 6K + 12K + 1. From the intersection of two straight lines y = KX + 2K + 1 and X + 2y-4 = 0 in the fourth quadrant, x = 2 − 4k2k + 1 ＞ 0y = 6K + 12K + 1 ＜ 0 can be obtained. Solving this inequality system, we can get − 12 ＜ K ＜ 16, that is, the value range of K is (− 12, − 16), so we choose C
If the two intersections of the straight line y = KX + 3 and the curve X ^ 2 + y ^ 2 + 2kx-3y-3 = 0 are symmetric about the Y axis, then the coordinates of the intersection are?
（±√3,3）.
If the two intersection points are symmetrical about the y-axis, then the diameter of the circle is the y-axis, that is, the center of the circle is on the y-axis, so k = 0. Thus, the linear equation is y = 3, the equation of the circle is X & sup2; + Y & sup2; - 3y-3 = 0, and the coordinates of the intersection point are (± √ 3,3)
If the two intersections of the line y = KX + 3 and the curve X square + y square + 2kx-3y-3 = 0 are symmetric about the Y axis, then the coordinates of the intersection are?
Because the intersection is symmetric about the Y axis, the line is perpendicular to the Y axis, that is, k = 0, and the equation of the line is y = 3,
Substituting into the curve equation, we can get X1 = - √ 3, X2 = √ 3,
Therefore, the coordinates of the intersection point are (- √ 3,3), (√ 3,3)
It is known that m and N are the common vertex of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 and hyperbola C2: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, P is the moving point on C2, and the slopes of lines MP, NP, MQ and NQ intersecting C1 at point Q (P and Q are different from m and N) are K1, K2, K3 and K4 respectively. It is proved that K1 + K2 + K3 + K4 is the fixed value
It is known that m and N are the common vertices of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 and hyperbola C2: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, P is the moving point on C2, and the line segment OP intersects C1 at point Q (points P and Q are different from points m and N)
1. If the coordinate of point P is (2,1), the eccentricity of C2 is 2, and the molecular root is 6, the equation of C1 is obtained;
2. Note that the slopes of MP, NP, MQ and NQ are K1, K2, K3 and K4 respectively, which proves that K1 + K2 + K3 + K4 is the fixed value
Prove: the tangent of hyperbola C1: x ^ 2-y ^ 2 = 5 and ellipse C2: 4x ^ 2 + 9y ^ 2 = 72 at the intersection are perpendicular to each other
On derivative
The intersection coordinates (3,2) (3, - 2) (- 3,2) (- 3,2) (- 3, - 2) of the two equations are 3x-2y = 5, 2x + 3Y = 123x + 2Y = 5, 2x-3y = 12-3x-2y = 5, - 2x + 3Y = 12-3x + 2Y = 5, - 2x-3y = 12. The product of the slope is - 1, so you must ask how to solve the tangent equation, there is a conclusion
RELATED INFORMATIONS
- 1. The left and right fixed points of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / 4 = 1 (a is greater than 2) are a and B respectively, and point P is the image of hyperbola C2: x ^ 2 / A ^ 2-y ^ 2 / 4 = 1 in the first quadrant If the area of triangle ACD and triangle PCD is equal, (1) calculate the coordinates of point p; (2) whether the line CD can pass the right focus of ellipse C1, if it can, calculate the value of A. if not, please explain the reason
- 2. If there are two different intersections between the line y = KX + 2 and the circle (X-2) + (Y-3) = 1, find the value range of K
- 3. If the line y = KX (k > 0) and hyperbola y = 4 / X intersect at two points a (x1, Y1) and B (X2, Y2), then the value of 2x1y2-7x2y2
- 4. Prove: no matter what the value of K is, the straight line L: kx-y-4k + 3 = 0 and the curve C: x ^ 2 + y ^ 2-6x-8y + 21 = 0 always have several intersections 1. To make the line intersect the circle, the distance from the center of the circle to the line is | 3k-4-4k + 3 | / (radical K ^ 2 + 1) Radical (1 + 2K / (1 + K ^ 2))
- 5. The position relationship between the square of X + the square of Y - 6x = 0 and the square of circle x + the square of Y - 8y = 0. If they intersect, the common chord length can be obtained,
- 6. It is known that the center of the hyperbola is at the origin, the focal points F 1 and F 2 are on the coordinate axis, the eccentricity is 2 ^ 1 / 2 and passes through the point (4, - 10 ^ 1 / 2) It is known that the center of the hyperbola is at the origin, the focus F 1 and F 2 are on the coordinate axis, the eccentricity is 2 ^ 1 / 2 and passes through the point (4, - 10 ^ 1 / 2). Find the hyperbolic equation; if the point m (3, M) is on the hyperbola, find out: the point m is on the circle with the diameter of F 1F 2; find out the area of the triangle f 1m F 2
- 7. The coordinates of the focus are (- 6,0), (6,0) and the standard equation of hyperbola passing through point a (- 5,2)
- 8. Some problems on the standard equation of hyperbola 1. The focal coordinates are (0, - 6), (0,6) and (2, - 5) respectively 2. Eccentricity e = 4 / 3, imaginary axis length is 2 times root 7, focus on Y axis 3. The focal length is 10, the distance between two vertices is 6, and the vertex is on the X axis
- 9. Given that K is a real number, if the focal length of hyperbola X2K − 5 + Y22 − | K | = 1 has nothing to do with the value of K, then the value range of K is () A. （-2，0]B. （-2，0）∪（0，2）C. [0，2）D. [-1，0）∪（0，2]
- 10. Let the center of the ellipse be at the origin o, the focus be on the x-axis, the eccentricity be √ 2 / 2, and the sum of the distances between a point P and two focus points on the ellipse be equal to √ 6. Let the center of the ellipse be at the origin o, the focus be on the x-axis, the eccentricity be √ 2 / 2, and the sum of the distances between a point P and two focus points on the ellipse be equal to √ 6. (1) if the straight line x + y + M = 0 intersects the ellipse at two points a and B, and OA ⊥ ob, find the value of M?
- 11. The left and right focus of the hyperbola x2a2 − y2b2 = 1 is F1, F2, P is a point on the hyperbola, satisfying | PF2 | = | F1F2 |, and the straight line Pf1 is tangent to the circle x2 + y2 = A2, then the eccentricity e of the hyperbola is () A. 3B. 233C. 53D. 54
- 12. Point P is a point on the hyperbola x2 / 25-y2 / 9 with F1 and F2 as the focus, and the absolute value of Pf1 = 12, then the absolute value of PF2=
- 13. It is known that the focus of the ellipse x ^ 2 / 4 + y ^ 2 = 1 is F1, F2, and the intersection of the parabola y ^ 2 = PX (P > 0) and the ellipse in the first quadrant is Q, if ∠ f1qf2 = 60 °, (1) Calculate the area of △ f1qf2; (2) Find the equation of this parabola
- 14. Given that - 90 degree is less than X and less than 0 degree, SiNx + cosx = 0.2, find SiNx cosx =?
- 15. F (x) = the maximum of SiNx cosx
- 16. Are there overlaps between infinite discontinuities and oscillatory discontinuities, such as piecewise function x < 0, f (x) = 1 / x, x > 0, f (x) = sin Is there any overlap between infinite discontinuity and oscillation discontinuity For example, when the piecewise function x < 0, f (x) = 1 / x, x > 0, f (x) = sin (1 / x), what is the breakpoint of x = 0?
- 17. F (x) = sin (ω x + φ) (ω is greater than 0, φ is greater than or equal to 0 and less than or equal to π) is even function, and the distance between two adjacent highest points on the image is 2 π It is known that the function f (x) = sin (ω x + φ) (ω greater than 0,0 ≤ φ ≤ π) is even function, and the distance between two adjacent highest points on the image is 2 π 1. Find the analytic expression of F (x) 2. If a belongs to (- π / 3, π / 2), f (a + (π / 3)) = 1 / 3, find the value of sin (2a + (5 π / 3))
- 18. If the sum of the first n terms of the equal ratio sequence {an} is 2N-1, then the sum of the first n terms of the sequence {an ^ 2} is?
- 19. Let the sum of the first n terms of the equal ratio sequence {an} be Sn, S4 = 1, S8 = 17, and find the general term formula an
- 20. Let the sum of the first n terms of the sequence an be Sn, and Sn = n ^ 2-4n + 4, BN = an / 2 ^ n. can we use dislocation subtraction to find the sum of the first n terms of BN and TN?