Solution inequality: (x + 3) 2 + (x + 2) (x-1) > (x-3) (2x + 1) - 2

Solution inequality: (x + 3) 2 + (x + 2) (x-1) > (x-3) (2x + 1) - 2

（x+3）2+（x+2）（x-1）＞（x-3）（2x+1）-2，x2+6x+9+x2+x-2＞2x2-5x-3-2，7x+5x＞-5+2-9，12x＞-12，x＞-1．

If the inequality 2x-1 ＞ m (the square of x-1) holds for all - 2 ＜ = m ＜ = 2 and all m, find the range of real number X

If we change the original formula, then we have (x squared-1) * m-2x + 1

5x-17 = 3, what is x equal to

5x-17=3 x
5x-3x=17
2x=17
x=17÷2
x=8.5

There is a point P (1,1) in the ellipse x23 + Y22 = 1. A straight line passes through the point P and intersects with the ellipse at P1 and P2. The chord p1p2 is bisected by the point P. the equation of the straight line p1p2 is obtained

Let P1 (x1, Y1) and P2 (X2, Y2), then x123 + y122 = 1 and x223 + y222 = 1 are subtracted to get (x1 + x2) (x1 − x2) 3 + (Y1 + Y2) (Y1 − Y2) 2 = 0 ∵ chord p1p2 is bisected by point P, ∵ X1 + x2 = 2, Y1 + Y22 = 2. Substituting into the above formula, Y1 − y2x1 − x2 = - 23 is obtained, that is, the slope of line p1p2 is 23 ∵ line p1p2

The representation of numerical value in computer is generally 8421BCD code, and the representation of 8421BCD code of decimal number 68 is 8421BCD code

8421BCD is actually binary for decimal, but every four bit binary code represents a decimal number
eight thousand four hundred and twenty-one
（68）D =（0110 1000）B

A math problem: - 1 ^ 2 * (- 3) ^ 2 - (- 1 / 2) ^ 2013 * (- 2) ^ 2012 △ 2 / 9
Seventh grade volume I Chapter II rational number operation content Oh, thank you!

=－1×9－﹙－½×﹙﹣2﹚﹚²º¹²×﹙﹣½﹚÷2/9
=﹣9＋½÷2/9
=﹣9＋9/4
=﹣27/4

Given the function f (x) = LG [A / (x ^ 2 + 1)], there exists x0 in the domain of definition such that f (x0 + 1) = f (x0) + F (1), find the value range of a? (^ 2 means square)

lg[a/｛（x0+1）^2+1)｝=lg[a/(x0^2+1)] + lg[a/(1+1)]
=lg｛a/(x0^2+1)] * [a/(1+1)] ｝
A / {(x0 + 1) ^ 2 + 1)} = A / (x0 ^ 2 + 1)] * [A / (1 + 1)] has real roots

Given a = 2009, B = 2010, C = 2011, find the value of a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac

b-a=2010-2009=1
c-a=2011-2009=2
c-b=2011-2010=1
a^2+b^2+c^2-ab-bc-ac
=1/2(b-a)^2+1/2(c-a)^2+1/2(c-b)^2
=1/2×1+1/2×4+1/2×1
=3

One jump P jumps from point a one unit away from the origin to the origin, and jumps to point A1 for the first time
(the distance between point A1 and origin o is half of the distance between point a and origin o), jump to A2 for the second time (the distance between point A2 and origin o is half of the distance between point A1 and origin o), and keep beating like this What is the distance from point P to the origin after the first jump? After the second jump? After the 10th beat?

The distance from a to the origin is 1
After the first jump, the distance from point P to the origin is 1 / 2
After the second jump, the distance from point P to the origin is 1 / 2 * 1 / 2 = 1 / 2 ^ 2 = 1 / 4
After the 10th jump, the distance from point P to the origin is 1 / 2 ^ 10 = 1 / 1024

Define a new operation a * b = a (B + 6), then equation 3 * x = 2 * (- 9)

The rule a * b = a (B + 6) is applied to 3 * x = 2 * (- 9)
3(x+6)=2(-9+6)
The results are as follows
3x+18=-6
x=-8