a|x-1|

a|x-1|

When a is less than or equal to 0, X can take any value
When a is greater than 0, | X-1|

Solving inequality a (x-a) > X-1

a(x-a)>x-1
ax-a^2-x+1>0
(a-1)x>a^2-1
When a > 1, x > A + 1
When a

The solution of inequality (x-a) / a (x-1) about X
＜0

It can be divided into a > 0, A0
a> The solution of 1 x is 1

Formula of distance between two straight lines
Parallel straight line give two straight line equation, find the distance

The two parallel lines are L1: ax + by + C1 = 0, L2: ax + by + C2 = 0. Take any point P (x0, Y0) in L2, then ax0 + by0 + C2 = 0, ax0 + by0 = - C2. According to the formula of distance from point to line, the distance from P to L1 is: | ax0 + by0 + C1 | / √ (A & # 178; + B & # 178;) = | - C2 + C1 | / √ (A & # 178; + B & # 178;) = | C1-C2 |

It is proved by the counter proof method that when P and Q are all odd numbers, the roots of the equation x ^ 2 + 2px + 2q = 0 (P ^ 2-2q is greater than 0) are irrational numbers
RT

Suppose two are m, N, then they must be integers
When m and N are both odd numbers, Mn is odd. It is contradictory to Mn = 2q
When m and N are even numbers, Mn is a multiple of 4, which is contradictory to Mn = 2q
When m and N are an odd number and an even number, M + n is an odd number
So: the equation x ^ 2 + 2px + 2q = 0 is irrational

The observation method is used to find the special solution of the differential equation d ^ 2Y / DX ^ 2-3 * dy / DX + 2Y = 5
D ^ 2Y / DX ^ 2-3 * dy / DX + 2Y = 5, how can we get the detailed process of 5 / 2 through observation

Because the right side is a constant, the first and second derivatives are both 0, which becomes 0 when substituted into the original equation
So what's left is 2Y = 5, so we get the special solution y = 5 / 2

What is the projective theorem, represented by a graph

In the formula RT △ ABC, ∠ BAC = 90 ° ad is the height on the hypotenuse BC, then there is a projective theorem as follows:
AD^2 = BD·DC
AB^2 = BD·BC
AC^2 = CD·BC
AB·AC = BC·AD

If the range of quadratic function f (x) = AX2 + 2x + C (x ∈ R) is [0, + ∞), then the minimum value of a + 1C + C + 1a is ()
A. 2B. 2+2C. 4D. 2+22

If f (x) is a quadratic function, then a ≠ 0. From the meaning of the problem, we can see that △ is equal to 0, and AC = 1. By using the properties of inequality, we can get a + 1C + C + 1A = A2 + C2 + A + CAC ≥ 2 + 2Ac ≥ 4, so we choose C

Find the derivative of the function y = ln (1 + 2 ^ x)

The derivative of y = ln (1 + 2 ^ x)
y'=[1/(1+2^x)]*(1+2^x)'
=[1/(1+2^x)]*(2^x)*ln2
=[(2^x)*ln2]/(1+2^x)

The fraction is 295. After adding m to numerator and denominator, the ratio of numerator to denominator is 19:7. What is m?

29 + M5 + M = 197, & nbsp; 19 × (5 + m) = 7 × (29 + m), & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 95 + 19m = 203 + 7m, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 12m = 108, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; m = 9