-5a+(3a-2)-(3a-7)=?

-5a+(3a-2)-(3a-7)=?

-5a+(3a-2)-(3a-7)
=-5a+3a-2-3a+7
=-5a+3a-3a-2+7
=-5a+5
=5-5a

Calculate - 5A + 2 (3a-2) - 2 (3a-7),

-5a+2（3a-2）-2（3a-7）
=-5a+6a-4-6a+14
=-5a+10

If the solution of the system of equations 3x-y = 7m + 3x + 2Y = 8 satisfies X-Y

3x-y = 7m + 36x-2y = 14m + 6x + 2Y = 8, 7x = 14m + 14x = 2m + 2
Substituting x + 2Y = 8 2m + 2 + 2Y = 8 y = 3-m
x+y=2m+2-(3-m)=3m-1

The solution of LG (5 ^ x + 2x-3) = x (1-lg2) is__________

lg(5^x+2x-3)=x(1-lg2)
lg(5^x+2x-3)=x(lg10-lg2)
lg(5^x+2x-3)=x(lg10/2)
lg(5^x+2x-3)=xlg5
lg(5^x+2x-3)=lg5^x
5^x+2x-3=5^x
2x=3
x=3/2

Inequality 1 / (x ^ 2 - 2x + 2) + LG {(1-x ^ 2 + 2x) / (1 + x ^ 2-2x)} < 1 / 2

Exchange yuan
Let t = x ^ 2-2x
Then the left side of the original inequality becomes 1 / (T + 2) + LG [(1-T) / (1 + T)]
The value range of T is (- 1,1)
1 / (T + 2) is monotonically decreasing
(1-T) / (1 + T) is also monotonically decreasing
So LG [(1-T) / (1 + T)] is also single decreasing
So 1 / (T + 2) + LG [(1-T) / (1 + T)] is also monotonically decreasing
When t = 0, 1 / (T + 2) + LG [(1-T) / (1 + T)] = 1 / 2
So 0

Seeking inequality x2 + 3x + 5

Let y = x & # 178; + 3x + 5
∵1＞0
The opening of the graph of functions with y = x & # 178; + 3x + 5 is upward
∵△＝3²－4×1×5
＝9－20
＝﹣11＜0
There is no intersection point between the function image of y = x & # 178; + 3x + 5 and X axis
Ψ y = x ^ 2 + 3x + 5 > 0 (x is any real number)
The inequality x ^ 2 + 3x + 5

A (x1, Y1) B (X2, Y2) C (X3, Y3) find the vectors AB, BC, CA, and verify the vector AB + BC + Ca = 0
The ab = -bc-ca (x2-x1, y2-y1-x1-x1-x1, and y2-y1-x1-x1 (x2-x1, x2-x1-x1, y2-y1-y1) and the ab = -bc-ca (((x2-x1-x1, x2-x1-x1-x1, and y2-y1-y1-x1-x1, which is the ab = -bc-bc-ca (((x2-x1-x1-x1, and y2-y1-y1-y1-y1-y1-y1))) & nbsp & & nbsp; & & nbsp; & & nbsp & nbsp; & & amp & nbsp; & & amp & nbsp; & & nbsp; & & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & & nbsp; & nbsp & nbsp; & nbsp; & nbsp; is the, the, the AB is such, the AB is not such, the, the AB is such, the AB, the AB is the AB, AB, AB, AB, AB, AB, AB, AB, AB, AB is the& nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; Whether or not the AB + BC (or AB + BC + Ca = (x2-x1, y2-x1, y2-y1) + (x3-x2, y3-y3, y3-y2) + (x3-x2, y3-y22, and (x1-x3, y1-y1-y3, y1-y3) is the next AB + BC (or AB + BC + BC + Ca = (x1-x3x3x3x3, y1-y3, y1-y3-y3-y3-y3) is not the AB + BC (or AB + BC + AB + BC + BC (or AB + BC) is the AB + BC + BC + BC (or (or (AB + BC) is the following (or AB + AB + BC) is the following (or AB + BC) or AB + BC (AB + BC + BC) is the (or AB + AB + BC (or AB + AB + BC) or AB + BC (AB + BC) is the (or AB + AB + BC (or AB + BC nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; &Nbsp; & nbsp; = (0,0) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;? Is that right?

Your method is too cumbersome
Vector AB = vector ob - vector OA = (x2-x1, y2-y1)
Vector BC = vector oc - vector ob = (x3-x2, y3-y2)
Vector CA = vector OA - vector OC = (x1-x3, y1-y3)
So vector AB + vector BC + vector CA = (0,0) = 0 vector

How to prove the formula of triangle center of gravity x = (x1 + x2 + x3) / 3, y = (Y1 + Y2 + Y3) / 3

The formula of the center of gravity of an inscribed triangle is proved in the coordinate system. It is proved that: take a point of a triangle as the origin, the center of gravity is the intersection of the vertical lines on the three sides of the triangle, and take the abscissa of the three points of the triangle as x1.x2.x3 respectively. Line x1x2 = line x2x3. So the abscissa of the center of gravity is x = X1 + x2 + X3. Similarly, the ordinate of the triangle is y = Y1 + Y2 + Y3

Let f (x) have f (x1 + x2) = f (x1) f (x2) for any real number x1, X2, and f '(0) = 1. It is proved that f' (x) = f (x)

It is proved that: (I) Let f (x) be non-zero in the domain of definition. From the original formula: | f (x + y) | = | f (x) | * | f (y) | thus: ln | f (x + y) | = ln | f (x) | + ln | f (y) | both sides of the equation derive y at the same time, and then: (x + y)'f '(x + y) / F (x + y) = f' (y) / F (y) + 0 is transferred and sorted: F '(x + y) = f (x + y) f' (y) / F (y) = f '(y) f (...)

Let f (x) have f (x1 + x2) = f (x1) f (x2) and f '(0) = 1 for any real number x1, X2, and prove that f' (x) = f (x)
It's written step by step
Back to the first floor
No idea

F (0 + 0) = f (0) f (0), that is, f (0) = f (0) ^ 2, then f (0) = 0 or 1 if f (0) = 0. Let x2 = 0, then f (x + 0) = f (x) f (0) = 0, that is, f (x) = 0 contradicts f '(0) = 1, so f (0) is not 0. F (0) = 1F (0) = f (x-x) = f (x) f (- x) = 1, so f (- x) = 1 / F (x) f' (x-x) = f '(x) f (- x) + F (x) f' (- x) = f '(x) / F (...)