In order to make the inequality x-2ax + 1 greater than or equal to half of the square of X-1 hold true for all real numbers x, find the solution of real number a

Double two on both sides
2x²-4ax+2>=x²-2x+1
X & # 178; + (2-4a) x + 1 > = 0
So Delta

The solution set of inequality 2x ^ 2 + MX + n > 0 is {x | x > 3 or X

On the inequality of X - 1 / 2x ^ 2 + 2x ≥ MX, the solution set is (0,2), find the value of M!

The inequality is arranged as follows
-x²+4x≥2mx
x²+(2m-4)x≤0
x[x+(2m-4)]0
So:
[x+(2m-4)]

If A1, A2, A3 are linearly correlated, then vector group B: A1, A2, A3, a1 + A2 ()
As long as the answer is OK

If A1, A2, A3 are linearly correlated, then vector group B: A1, A2, A3, a1 + A2 (linearly correlated,)

This program is to use dichotomy to find the root of equation x to the fourth power - 3x + 1 in the interval of 0.3 to 0.4, and the error should not exceed the negative second power of 0.2 × 10
The sooner the better!

#include
#include
#define F(y) (y*y*y*y-3.0*y+1.0)
main()
{
double a,b,a1,b1,x;
int k;
double t=0.2e-2;
a=0.3;
b=0.4;
j=log(200*(b-a))/log(2);
n=(int)j
a1=(F(a)>F(b)?a:b);
b1=(F(a)

It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, A2 = 4, S5 = 35. (I) find the sum of the first n terms of the sequence {an}; (II) if the sequence {BN} satisfies BN
It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, A2 = 4, S5 = 35. (I) find the sum of the first n terms of the sequence {an} and Sn; (II) if the sequence {BN} satisfies BN = ean, find the sum of the first n terms of the sequence {BN} and TN

1.a2=a1+d=4
Sn = Na1 + n (n-1) d / 2, that is, S5 = 5A1 + 10d = 35
The solution is A1 = 1, d = 3  Sn = n + 3N (n-1) / 2 = n (3n-1) / 2
2.bn=e^an=e^(3n-2)
b(n+1)=e^(3n+1)
b(n+1)/bn=e^3
So BN is an equal ratio sequence, the first term B1 = e, and the common ratio is e ^ 3
Tn=e(1-e^3n)/(1-e^3)

1. For the equation of X, the solution of x2 + 2a-3 + A2 = 0 is -- 2. If x2-6x + m is a complete square, then M = -- (M is a constant)

1 x2=a2-2a+3
X is equal to the positive and negative root sign a2-2a + 3
2 x2-6x+m
m=9
So, 6 is the coefficient of x2 times 2 times the root M
So the root sign M = 3
So m = 9

The coordinates of △ ABC vertex are a (3,4) B (0,0) C (m, 0) 1. When m = 5, the value of sina is 2. When a is an obtuse angle, the range of M is 2

1. When m = 5, ab = 5, BC = 5, AC = 2 √ 5; from the cosine theorem: cosa = (AC ^ 2 + AB ^ 2-bc ^ 2) / (2Ac * AB) = (20 + 25-25) / (2 * 5 * 2 √ 5) = √ 5 / 5, so: Sina = 2 √ 5 / 5.2

When x, the zeroth power of (x-4) = 1

X≠4

Among the points representing integers (except the origin) on the number axis, the point closest to the origin has The number is ．

As shown in the figure: it can be seen from the figure that there are two points closest to the origin, namely ± 1. So the answer is: 2, ± 1

In order to make the inequality x-2ax + 1 greater than or equal to half of the square of X-1 hold true for all real numbers x, find the solution of real number a