Square of (2x + 1) = 3 (2x + 1)

Square of (2x + 1) = 3 (2x + 1)

Classmate, this problem can be solved in this way
4x^+4x+1-6x-3=0
4x^-2x-2=0
(4x+2)(x-1)=0
So x = - 1 / 2 or 1
The cross multiplication is mainly used
Of course, you can also use the square difference formula
Square of (2x + 1) = 3 (2x + 1)
The square of (2x + 1) - 3 (2x + 1) = 0
（2x+1）（2x+1-3）=0
（2x+1）（2x-2）=0
X = - 0.5, or x = 1
These are the two solutions I've provided,

Square of (x + 2x) - 3 (x + 2x) + 2 = 0

The square of (the square of the square of + 2x) is the square of (the square of the square of + 2x (the square of the square of + 2XX + 2x + 2x) of the square of (the square of the square of the square of + 2x + 2x) of the square of (the square of the square of + 2x + 2x + 2x) of (x + 2x + 2x) + 2 = 0 [(x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x = - 1 ± √ 2 or x = -

It is known that a (1, - 2) B (2,1) C (3,2) and D (- 2,3) are represented by vector ab. vector AC is used to represent vector AD + vector BD + vector CD

AB=（1,3）
AC=（2,4）
AD=（-3,5）
BD=（-4,2）
CD=（-5,1）
AD+BD+CD=（-12,8）
With X, y
x*AB+y*AC=AD+BD+CD
x+2*y=-12
3*x+4*y=8
x=32；y=-22
So 32 * ab-22 * AC = AD + BD + CD

In the triangle ABC, D is the midpoint of AB and E is the midpoint of AC. it is known that the area of the triangle ade is 8 square centimeters, and the area of the shadow part is calculated
The shaded part is the triangle bec

Because D is the midpoint of AB and E is the midpoint of AC,
So De is the median of △ ABC
So s Delta ABC:S △BEC=4：1
So s △ ABC = 32 square centimeters
And because e is the midpoint of AC
So s △ BEC = 1 / 2 * s △ ABC = 16 square centimeters

It is known that f 1F 2 is the two focal points of the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1, P is a point in the second quadrant of the ellipse, ∠ pf1f 2 = 120 ° and the area of △ pf1f 2 is calculated

From the meaning of the title: a = 2, B = 3, C = 1,
In △ pf1f2, the ellipse defines Pf1 + PF2 = 2A = 4
Let Pf1 = t and ∵ pf1f2 = 120 ° be obtained from cosine theorem
（4－T）＾2＝T＾2＋4－2＊T＊2＊sin120°
The solution is t = 6 / 5, let P (m, H), then H = t * sin 60 ° = (3 √ 3) / 5
S △PF1F2＝1／2 ＊（F1F2）＊h＝（3√3）／5
The area of 〈△ pf1f2 is (3 √ 3) / 5

Finding the inscribed circle radius of right triangle
The three sides of a right triangle are 3, 4 and 5. Find the radius of the inscribed circle of the right triangle,

Let R be the radius of the inscribed circle,
From the tangent length theorem, we can get: 3-R = 5 - (4-R), so r = (4 + 3-5) / 2 = 1

Given the function f (x) = e ^ x, G (x) = KX, X belongs to R
1. If k = e ^ 2, try to determine the monotone interval of F (x) - G (x)
2. If k > 0 and f (LXL) > G (LXL) is constant for any x, try to determine the value range of real number K

1. Let H (x) = F-G = e ^ x-xe ^ 2
h'(x)=e^x-e^2
When x > 2, H '(x) > 0, monotonically increasing
When x

Try to explain that the sum of the two sides of a right triangle is equal to the sum of its circumcircle diameter and inscribed circle diameter
A good answer is rewarded

The diameter of right triangle inscribed circle is a + b-c
The circumcircle diameter of right triangle is: C
So there are the above conclusions

The known set a = {x | x ^ 2-x-2 > = 0}, B = {x | x + a}|

This is the CUA = {{x {x {x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\realr

In the quadrilateral ABCD, ab = BC = CD = Da = BD = 1, the value range of AC is______

In the quadrilateral ABCD, it becomes a space tetrahedron. When a is close to the coincidence of C, the distance of AC is close to 0; when ABCD is close to a plane figure, the maximum distance of AC is close to 3, so AC ∈ (0,3), so the answer is: (0,3)