(x minus 1) = (x minus 4) = (x minus 5) = (x minus 1) = (x minus 2) = (x minus 4) = (x minus 5) 1

(x minus 1) = (x minus 4) = (x minus 5) = (x minus 1) = (x minus 2) = (x minus 4) = (x minus 5) 1

(x-4)(x-5)=（x-1)(x-2)
x^2-9x+20=x^2-3x+2
9x-3x=20-2
6x=18
x=3
Hope it can help you, please accept

The problem of setting unit of unknowns in solving practical problems
Let's set the unknown number as X, how can we bring the unit? For example, in the title, we say that the efficiency of Party B is 25% of that of Party A. I want to set the efficiency of Party A as X, what can we say about the unit after X, or we don't need to bring the unit when we set the efficiency?

In these cases, the unit is not the key point. The unit is just a scalar quantity. Here, X is the positioning unit, and then the following is unified as one

If the points a (3,5, - 7) and B (- 2,4,3) are known, the projection length of line AB on the coordinate plane YOZ is
Given the points a (3,5, - 7) and B (- 2,4,3), the length of the projection of line AB on the coordinate plane YOZ is

|The length of projection | ^ 2 = (5-4) ^ 2 + (3 + 7) ^ 2 = 101
The length of the projection of line AB on the coordinate plane YOZ is the root of 101

In the right triangle ABC, ∠ a = 90 °, the opposite sides of BC, AC and ab are a, B, C, and / A-41 / + (B-40) &# 178; = 0, then what is C?

/A-41 / + (B-40) & #178; = 0, then: A-41 / = 0, (B-40) & #178; = 0. So: a = 41, B = 40
C & # 178; = A & # 178; - B & # 178; = 41 & # 178; - 40 & # 178; = 81. So C = 9

If the two roots of the equation 2x square + (K + 1) X-5 = 0 are opposite to each other, what is the value of K

Two roots a and B ∵ A and B are opposite numbers ∵ a + B = 0
And a + B = - (K + 1) / 2 (K + 1) / 2 = 0
∴k=1

Points a and B are the left and right ends of the major axis of the ellipse x ^ 2 / 36 + y ^ 2 / 20 = 1 respectively. Point F is the right focus of the ellipse. Point P is on the ellipse and above the X axis. Ba is perpendicular to PF
(1) Finding point P coordinate
(2) Let m be a point on the major axis ab of the ellipse, and the distance between M and the straight line BP is equal to | Ma |, then the minimum distance d from the point on the ellipse to the point m is obtained

1. Calculate the coordinates of a, B and F first
A(-6,0),B(6,0),F(4,0)
Point P coordinate, set as (x, y), Y > 0, x ^ 2 / 36 + y ^ 2 / 20 = 1 (1)
PA and PF are perpendicular to each other, and their slopes are multiplied by - 1
That is, Y / (x + 6) * y / (x-4) = - 1 = > y ^ 2 = - (x + 6) (x-4) (2)
Substituting (2) into (1), x ^ 2 / 36 - (x + 6) (x-4) / 20 = 1
=> 2x^2+9x-18=0
=>X = 3 / 2, x = - 6 in (2)
When x = 3 / 2, y ^ 2 = - (6 + 1.5) / (1.5-4) = 7.5 / 2.5 = 3 and Y > 0, so y = √ 3
When x = - 6, y ^ 2 = 0 = > y = 0 is out of the question
So the coordinates of the point P are (3 / 2, √ 3)
2, from P (3 / 2,5 √ 3 / 2), we get l (AP): (y-0) / (5 √ 3 / 2-0) = (x + 6) / (3 / 2 + 6), then l (AP): X - √ 3Y + 6 = 0 ∵ m to AP distance = lmbl, m (x, 0) | x + 6 | / 2 = L6 XL (- 6)

In the known triangle ABC, angle a is equal to 90 degrees, ab = AC, D is the midpoint of BC, e and F are the points on ABAC, be = AF, and the triangle BDE and triangle CDF are calculated
Angle BDE relations, urgent, online, etc

There is no specific relationship between s △ BDE and △ CDF through the existing conditions. If be = CF is added, then △ BDE = △ CDF can be determined;
In terms of area, the area of △ BDE + △ CDF is equal to half of △ ABC;
In addition, several congruent triangles △ BDE = △ ADF, △ AED = △ CFD can be determined

How to solve the problem that two times of the fourth power of x minus the square of x equals 28,
I want to ask you how to go from the second step to the third step and how to think. I understand from the third step to the second step. I just don't understand from the second step to the third step

2x^4-x^2=28
(2x^2+7)(x^2-4)=0
x^2=-7/2 or x^2=4
x=2,-2,i√(7/2),-i√(7/2)

Find the function f (x) = log2 (a ^ X - (2 ^ x) * k), a is greater than or equal to 2, and K is a constant, find the domain of the function
The teacher seems to be talking about a,

a^x-(2^x)*k>0
a^x>(2^x)*k
(1)a>2
When K0
xlga>xlg2+lgk
Domain x > LGK / (lga-lg2)
(2) a=2
When k = 1, a ^ x > (2 ^ x) * k does not hold, and the domain is an empty set,
It doesn't match the function definition. It's impossible

If the length of a rectangle increases by 4cm and the width decreases by 1cm, the area remains unchanged; if the length decreases by 2cm and the width increases by 1cm, the area remains unchanged, the area of the rectangle is______ ．

Let the length of the rectangle be xcm and the width be YCM. According to the meaning of the question, we get (x + 4) (Y − 1) = XY (x − 2) (y + 1) = XY. The solution is x = 8y = 3, so xy = 8 × 3 = 24. Answer: the area of the rectangle is 24cm2. So the answer is 24cm2