The solution set of quadratic inequality KX ^ 2 + KX + 4 ≥ 0 with one variable is r, and the value range of real number k is obtained

The solution set of quadratic inequality KX ^ 2 + KX + 4 ≥ 0 with one variable is r, and the value range of real number k is obtained

The solution set of inequality is r
k> 0, and △ = k ^ 2-16k ≤ 0,
Yes, 0
The inequality | X-1 | ≤ 2 has the same solution set as the inequality AX2 + bx-2 ≤ 0. Find the value of real numbers a and B
-2≤x-1≤2
-1≤x≤3
Ax & # 178; + bx-2 ≤ 0, the solution set is - 1 ≤ x ≤ 3
Then - 1 and 3 are solutions of the equation AX & # 178; + bx-2 = 0
By Weida theorem
-1+3=-b/a
-1*3=-2/a
therefore
a=2/3
b=-4/3
If the solution set of inequality | x + 1 | - | X-2 | > k is r, then the value range of real number k is______ .
According to the absolute value inequality, we can get: | x + 1 | - | X-2 | ≤| (x + 1) - (X-2) | = 3, that is: - 3 ≤| x + 1 | - | X-2 | ≤ 3, so to satisfy | x + 1 | - | X-2 | > K, the solution set is r, only K ＜ - 3 is needed, so the answer is: (- ∞, - 3)
If the solution set of inequality KX ^ 2-2x + 6K < 0 about X is an empty set, the value range of real number k is obtained
If the solution set is empty, then KX & # 178; - 2x + 6K ≥ 0 holds
K=0
-2x ≥ 0, not constant
K is not equal to 0
If the quadratic function is always greater than or equal to 0, the opening is upward
K>0
And the discriminant is less than or equal to 0
So 4-24k and 178; ≤ 0
k²≥1/6
K>0
So K ≥ √ 6 / 6
ditto
kx^2-2x+6k＜0
case 1: for k >0
Discriminant = 4-24k ^ 2 1 / 6
k> √6/6 or k0 and (k> √6/6 or k √6/6
case 2: k √6/6
It is known that in △ ABC, ad is the midline on the edge of BC, e is the point on ad, and be = AC, extend be to AC to F, and prove that AF = EF
Prove: as shown in the figure, extend ad to point G, so that ad = DG, connect BG. ∵ ad is the midline (known) on the edge of BC, ∵ DC = dB, in △ ADC and △ GDB, ad = DG ≌ ADC ≌ GDB (SAS), DC = DB ≌ ADC ≌ GDB (SAS), ≌ CAD = g, BG = AC and ∵ be = AC, ≌ be = BG, ≌
What is the coefficient of x ^ (- 2) in the expansion of binomial (x + (1 / x) - 2) ^ 5?
In the expansion (- 2,2) ^ (x) + (- 2) ^ (x) + - 2 * (- 2) ^ (x) + - 2) ^ (x) + - 2 * (- 5) ^ (x) + - 2) ^ (- 5) ^ (- 5) * (- 5) ^ (x) + - 2) ^ (- 5) ^ (x) + - 2] ^ (- 5) ^ (- 5) * (- 5) ^ (x) + - 2) ^ (x) ^ (- 5) ^ (- 5) ^ (x) + - 2 * (- 2) ^ (- 5) ^ (x) + - 2 * (- 2) ^ (- 5) ^ (- 5) ^ (x) ^ (x) + - 2] ^ (- 2) ^ (- 5) ^ (- 5) ^ (x * (- 2) ^ (- 5) ^ (x) + - 2) ^ (- 5) ^ (x * (- 2) ^ (- 5) ^ (- 5) ^ (x] ^ (- 5) ^
In the expansion, the coefficient of x ^ (- 2) is C (5,2) * - 2) ^ 3 + C (5,3) C (2,1) (- 2) = - 120
It is known that: as shown in the figure, in the plane rectangular coordinate system, the quadrilateral ABCO is a diamond, and ∠ AOC = 60 °, the coordinate of point B is (0,8 times the root 3), point P starts from point C and moves to point B on the line segment CB at the speed of 1 unit length per second, at the same time, point Q starts from point O and moves along the direction of ray OA at the speed of a (1 ≤ a ≤ 3) unit length per second. After t (0 < T ≤ 8) seconds, the line PQ intersects ob at point D
(1) Calculate the degree of ∠ AOB and the length of line OA;
(2) Find the analytical formula of the parabola passing through three points a, B and C;
(3) When a = 3, OD = 4 / 3 times the root 3, the value of T and the analytical expression of PQ of the straight line are obtained;
(4) When a is what value, the triangle with O, P, Q, D as vertex is similar to △ OAB? When a is what value, the triangle with O, P, Q, D as vertex is not similar to △ OAB? Please give your conclusion and prove it
Solution: (1) because the quadrilateral ABCO is rhombic, ∠ AOC = 60 & # 186;, so, ∠ AOB = 30 & # 186;. Connect AC to ob at m, then om = 1 / 2 × ob, am ⊥ ob. So am = tan30 & # 186; × om = 4. So OA = am / sin30 & # 186; = 8. (2) from (1), we can know a (4,4), B (0,8), C (- 4,4)
(1) Because ∠ AOC = 60 °, so ∠ AOB = 120 °, OA = ob = 8 × root 3;
(2) The analytical formula of parabola passing through points a, B and C is: y = - 1 / 12x & # 178; + 16?
Prove: let a be a square matrix of order n, if AX = 0 for any n-dimensional vector x, then a = 0
Such as the title
On the relationship between the two
(rank of a) + (dimension of solution space of AX = 0) = n
Now, according to the meaning of the problem, the solution space of AX = 0 is the whole space
(dimension of solution space AX = 0) = n
So the rank of a is zero, so a = 0
In the second proof, if a ≠ 0, then an element of a a (I, J) ≠ 0, let X be an n-ary sequence with the j-th component of 1 and the rest of the elements of 0, then the i-th component of the n-ary sequence ax is a (I, J) ≠ 0, which is contradictory to the proposition
It is known that Mn is the median line of triangle ABC, P is on Mn, BP and CP intersect on D, e. to prove AE: be + ad: DC = 1
Extend AP to BC in F, then to FG ∥ CE to AB in G, to FH ∥ BD to AC in H. ∵ Mn is the median line of AB and AC in △ ABC, ∥ Mn ∥ BC, ∥ MP ∥ BF, ∥ AP = PF. ∥ FG ∥ CE, AP = PF, ∥ AE = eg. ∥ FH ∥ BD, AP = PF, ∥ ad = DH
In the expansion of binomial (x2 − ax) 5, if the coefficient of X is - 10, then the value of real number a is -______ .
TR + 1 = Cr5 (x2) 5 − R (− ax) r = (− a) rcr5x10 − 2R − R, let 10-3r = 1 {r = 3, therefore, (- a) 3c53 = - 10} - A3 = - 10} a = 1