Given that the solution of inequality 3 + K (x + 2) - 4x ＞ K (x + 3) about X is negative, then the value range of K is negative

Given that the solution of inequality 3 + K (x + 2) - 4x ＞ K (x + 3) about X is negative, then the value range of K is negative

From 2x-1 > x + 1, we can get x > 2, from K (x + 3) > 3x + 14, we can get (K-3) x > 14-3k. (Note: therefore, we should discuss it in different cases) (1) when k = 3, the inequality does not hold. (2) when k is not equal to 3, we can get x > (14-3k) / (K-3) from (K-3) x > 14-3k
kk(X+3) =》 3+kX+2k-4X>KX+3k =》 4X
It is known that there are four integer solutions of the inequality system x-k ≥ 0.3-2x > - 1 about X, then what is the value range of K?
Can you explain why - 3 and - 2
Why - 3
x-k≥0 x≥k
There are four integer solutions of 3-2x > - 1 X-1, and the integers less than 2 are 1,0, - 1, - 2
So - 3
x-k≥0 x≥k
There are four integer solutions of 3-2x > - 1 X-1, and the integers less than 2 are 1,0, - 1, - 2
So - 3
It is known that there are five integer solutions of the inequality system x-k > = 0, 5-2x > 1 about X, then the value range of K is
x-k>=0
=> x>=k
5-2x>1
=> x
Given that x = 1 is the solution of the inequality k2x2-6kx + 8 ≥ 0 (K ≠ 0), then the value range of K is______ .
Because x = 1 is the solution of inequality k2x2-6kx + 8 ≥ 0 (K ≠ 0), so k2-6k + 8 ≥ 0, the solution is k ≥ 4 or K ≤ 2 and K ≠ 0. So the answer is k ≥ 4 or K ≤ 2 and K ≠ 0
As shown in the figure, EF is the median line of triangle ABC, and the bisector of angle ACB intersects EF at point D
prove:
∵ EF is the median of ⊿ ABC
∴EF//BC
∴∠FDC=∠BCD
∵∠ BCD = ∠ FCD [CE bisection ∠ ACB]
∴∠FDC=∠FCD
∴DF=FC
∵AF=CF
∴DF=AF
∴∠FAD=∠FDA
∴∠FCD+∠FAD=∠FDC+∠FDA=∠ADC
∵∠FCD+∠FAD+∠ADC=180º
∴∠ADC=90º
That is ad ⊥ CD
What is the maximum value of the binomial coefficients of (a + b) ^ n
When n is even, C (n, [n / 2]) is the largest;
When n is odd, C (n, [n + 1 / 2]) or C (n, [n-1 / 2]) is the largest
Move point a (1, - 3) to point A1 (3,0), and move point P (2,3) to point P1 according to the same translation method, then the coordinates of point P1 are（
P1(4l,6)
Known A1 ^ 2 + A2 ^ 2 + a3 ^ 2 + +an^2=1,x1^2+x2^2+x3^2+………… Xn ^ 2 = 1
a1 X1+a2 X2+a3 X3+………… An xn ≤ 1, using three theorems of basic inequality
a1^2+a2^2+a3^2+………… +an^2+1^2+x2^2+x3^2+………… xn^2-2(a1 X1+a2 X2+a3 X3+………… An xn) = (a1-x1) ^ 2 + (a2-x2) ^ 2 +. + (an xn) ^ 2 > = 0, so 2 (A1 X1 + A2 x2 + a3 X3 +...) an Xn)
In triangle ABC, EF is the median line of △ ABC, D is a point on the side of BC (not coincident with B and C), ad and EF intersect at point O, connecting de and DF. To make quadrilateral AEDF a parallelogram, conditions need to be added_______ (only add one), and write the proof process reference answer: (1) eo = of or
(2) Ed ‖ AC or
(3) BD = DC, or
I already know how to prove (3) condition, but (1), (2) I don't know how to prove it
（2）
∵ EF is the median of △ ABC
∴EF//BC
‖ angle B = angle AEF
∵ED‖AC
‖ angle AEF = angle B
All △ AEF is equal to △ EBD (ASA)
∴BD=EF
The midpoint of BC is d
∴DF//AB
The quadrangle AEDF is a parallelogram
(1)
I think
EF is the median of △ ABC
It should be certain
AO=DO
The diagonals are divided equally
The quadrangle AEDF is a parallelogram
How to find the coefficient of x ^ 3 in the expansion of binomial
For example, to find the constant term TM + 1 of the sixth power of (2 radical x minus 1 / radical x) = C (6, K) times the square of (2x), why is it equal to the K power of (- 1) times C (6, K) times the 6-k power of 2 times the 6-k / 2 power of X, and then times the - K / 2 power of X? I have taught myself mathematics.
Exception in the expansion of (x-radical 3) to the 10th power, what is the coefficient of X to the 6th power? Why don't the powers of X divide by 2 as before?
Write the general term of the binomial expansion,
Let the exponent of X be 3,
The constant before the letter is the coefficient of x ^ 3,