(x-3) &# 178; = 9-x & # 178; how do you do it

(x-3) &# 178; = 9-x & # 178; how do you do it

(x-3) &# 178; = 9-x & # 178; (x-3) &# 178; - (X & # 178; - 9) = 0 (x-3) &# 178; - (x + 3) (x-3) = 0 (x-3) (x-3-x-3) = 0 (x-3) x (- 6) = 0, x = 3, benefactor, I see that you are a unique talent in the Wulin with strange bones, magnificent appearance and wisdom
It is known that the parabola y = x & sup2; - (M & sup2; + 4) x-2m & sup2; - 12. When we try to find out the real value of M, the distance between the two intersections of the image and the x-axis is the smallest? The smallest
X & sup2; - (M & sup2; + 4) x-2m & sup2; - 12 = 0 of two a, B = > A + B = M & sup2; + 4; ab = - 2m & sup2; - 12 = > (a, 0); (B, 0) is a parabola y = x & sup2; - (M & sup2; + 4) x-2m & sup2; - 12 image and X axis of two intersections, the distance between the two intersections = | A-B | = D; D & sup2; = (a + b) & sup2; -
Two a, B = > A + B = M & sup2; + 4; ab = - 2m & sup2; - 12 of X & sup2; - (M & sup2; + 4) x-2m & sup2; - 12 = 0
=>(a, 0); (B, 0) are two intersections of parabola y = x & sup2; - (M & sup2; + 4) x-2m & sup2; - 12 image and X axis
The distance between the two intersections = | A-B | = D; D & sup2; = (a + b) & sup2; - 4AB = m ^ 4 + 8m & sup
Two a, B = > A + B = M & sup2; + 4; ab = - 2m & sup2; - 12 of X & sup2; - (M & sup2; + 4) x-2m & sup2; - 12 = 0
=>(a, 0); (B, 0) are two intersections of parabola y = x & sup2; - (M & sup2; + 4) x-2m & sup2; - 12 image and X axis
The distance between the two intersections = | A-B | = D; D & sup2; = (a + b) & sup2; - 4AB = m ^ 4 + 8m & sup2; + 16 + 8m & sup2; + 48 = m ^ 4 + 16m & sup2; + 64 = (M & sup2; + 8) & sup2;
=>D = M & sup2; + 8 > = 8 min (d) = 8... Ans when m = 0
(- 1 / 2Y + 2x) (3 / 2x-2 / 3Y + 1) calculation
(-1/2y+2x)(3/2x-2/3y+1)
=-3/4xy+3x²+1/3y²-4/3xy-1/2y+2x
=3x²+1/3y²+2x-1/2y-25/12xy
If the function f (x) = X3 + AX2 + BX + A2 has an extreme value of 10 when x = 1, then the value of a is ()
A. - 3 or 4B. 4C. - 3D. 3 or 4
The derivative function f (x) = X3 + AX2 + BX + A2 has an extreme value of 10 when x = 1, f ′ (1) = 2A + B + 3 = 0, f (1) = A2 + A + B + 1 = 10, a = - 3, B = 3 or a = 4, B = - 11. When a = - 3, f ′ (x) = 3x2-6x + 3 = 3 (x-1) 2 ≥ 0, x = 1 is not an extreme point. When a = 4, B = - 11, f ′ (x) = 3x2 + 8x-11 = (x-1) (3x + 11), the sign of derivative is changed around x = 1 A = 4, so B
Given that X1 is the root of the equation x + lgx = 2012 and X2 is the equation x + 10 ^ x + 2012, find the value of X1 + x2
Sorry, there is no wealth
x+lgx=2012 ------> lgx=2012-x
x+10^x=2012 ----> 10^x=2012-x
Draw the image of y = lgx, y = 10 ^ x, y = 2012-x in the same coordinate system
X1 and X2 are the abscissa of intersection a and B of y = 2012-x and y = lgx, y = 10 ^ x, respectively
The image of y = lgx, y = 10 ^ x is symmetric with respect to y = X
Y = 2012-x is perpendicular to y = x, and the intersection is m (10061006)
So a, B are symmetric with respect to M
∴ x1+x2=2*1006=2012
Given the parabola y = x2 + (2m + 1) x + m + 1, the value of M is obtained according to the following conditions: (1) if the parabola crosses the origin; (2) if the vertex of the parabola is on the x-axis; (3) if the symmetry axis of the parabola is x = 1
(1) ∵ the parabola y = x2 + (2m + 1) x + m + 1 passes through the origin, the ∵ point O (0, 0) satisfies the parabolic equation, ∵ 0 = m + 1, the solution is m = - 1; (2) ∵ the vertex of the parabola is on the x-axis, ∵ Δ= (2m + 1) 2-4 (M + 1) = 0, that is, 4m2-3 = 0, the solution is m = ± 32; (3) ∵ the symmetry axis of the parabola is x = 1, ∵ 2m + 1 = - 2, the solution is m = - 32
(1) 5 (2x-7y) - 3 (4x-10y) (2) 2x - [2 (x + 3Y) - 3 (x-2y)]
(1) The original formula = 10x-35y-12x + 30y = - 2x-5y; (2) the original formula = 2x-2x-6y + 3x-6y = 3x-12y
Given that the function f (x) = x & # 179; + ax & # 178; + BX + C obtains the extremum when x = - 2 / 3 and x = 1, the value of a, B and the monotone interval of the function are obtained
Because f (x) has extremum when x = - 2 / 3 and x = 1, f '(- 2 / 3) = 0, f' (1) = 0
The solution is a = 1 / 2, B = - 2
So f '(x) = 3x ^ 2-x-2, when x1, f (x) increases monotonically, otherwise, it decreases
Given that x 1 is the root of the equation x + lgx = 2 and x 2 is the root of the equation x + 10 ^ x = 2, find the value of x 1 + x 2
X + lgx = 2, it is judged to be monotone, and the left is monotone increasing function,
Therefore, there can only be one root; similarly, equation 2 has only one root; if X1 is the root of equation 1, then lgx1 is the root of equation 2 (substituting it), then lgx1 = x2;
So if we substitute X1 + lgx1 = 2, we have X1 + x2 = 2
Given the parabola y = x & sup2; + 2m + M & sup2; - 1 / 2m-2 / 3 (1), when m is any real number, can the intersection of the parabola be on the straight line y = 1 / 2 x - 3 / 2?
(2) If the line y = 1 / 2 x + m has no intersection with the parabola I, the value range of M is obtained
Give reasons
1. For the same X, the value of Y corresponding to the parabola should be greater than the value of Y corresponding to the straight line, so as to ensure that the point on the parabola is above the straight line
X & sup2; + 2m + M & sup2; - 1 / 2m-2 / 3 - 1 / 2 x - 3 / 2 > 0 holds for any x constant, that is, the y value corresponding to the equation Δ line corresponding to the inequality