As shown in the figure, C and D are two points on the line ab. given BC = 14ab, ad = 13ab, ab = 12cm, find the length of CD and BD

As shown in the figure, C and D are two points on the line ab. given BC = 14ab, ad = 13ab, ab = 12cm, find the length of CD and BD

∵ BC = 14ab, ad = 13ab, ab = 12cm, ∵ BC = 3cm, ad = 4cm, ∵ CD = ab-ad-bc = 12cm-4cm-3cm = 5cm, BD = BC + CD = 3cm + 5cm = 8cm

As shown in the figure, C and D are two points on the line ab. given BC = 14ab, ad = 13ab, ab = 12cm, find the length of CD and BD

∵ BC = 14ab, ad = 13ab, ab = 12cm, ∵ BC = 3cm, ad = 4cm, ∵ CD = ab-ad-bc = 12cm-4cm-3cm = 5cm, BD = BC + CD = 3cm + 5cm = 8cm

As shown in the figure, △ ABC is inscribed in ⊙ o, and ab = AC, diameter ad intersects BC at e, and F is the midpoint of OE. If BD / / CF, BC = 2, then the length of segment CD is
This is the last question of the last question,

According to the meaning of the title, △ ABC is inscribed in ⊙ o, ab = AC, diameter ad intersects BC in E, and F is the midpoint of OE
The point E is the midpoint of BC and ad ⊥ BC
BD / / CF
The bdcf of quadrilateral is rhombic
∴DE=EF
Let de = x connect ob
Then in △ BOE, OB = 3x, OE = 2x
According to the arcing string theorem
BE＾2+OE＾2=0B＾2
That is: 1 ＾ 2 + (2x) ＾ 2 = (3x) ＾ 2
The solution is x = √ 5 / 5
In △ CDE,
CD＾2=CE＾2+DE＾2
CD=√（CE＾2+DE＾2)
=√（√5/5）＾2+1＾2）
=√1.2
complete.

If three unequal real numbers a, 1 and B form an arithmetic sequence and A2, 1 and B2 form an arithmetic sequence, then 1A + 1b=______ ．

∵ a, 1, B are equal difference sequence, ∵ 2 = a + B (1) and ∵ A2, 1, B2 are equal ratio sequence, ∵ 1 = a2b2, ∵ three unequal real numbers a, 1, B, ∵ 1 = AB (2), and ∵ 1 ∵ 2 get 1A + 1B = 2

If three unequal real numbers a, 1 and B form an arithmetic sequence, and a ^ 2, 1 and B ^ 2 form an arithmetic sequence, then 1 / A + 1 / b=
The answer to this question is - 2, but I calculate ± 2. Why doesn't 2 hold?

There are four real numbers. The first three numbers form an equal ratio sequence, and their product is 216. The last three numbers form an equal difference sequence, and their sum is 12

Let the four real numbers be a / Q, a, AQ, B. if the first three numbers are equal, then a ≠ 0, Q ≠ 0
(a/q)×a×(aq)=216
a³=216
a=6
If the last three numbers are equal, then 2aq = a + B
A = 6, 12q = B + 6
b=12q-6
The sum of the last three numbers is 12, AQ + A + B = 12
6q+6+b=12
6q+b=6
B = 12q-6
6q+12q-6=6
18q=12
q=2/3
b=12q-6=12×(2/3) -6=2
a/q=6/(2/3)=9 aq=6×(2/3)=4
The four numbers are 9, 6, 4 and 2

It is known that three real numbers form an equal ratio sequence, and their product is 64. If the middle number is added with 1, it will form an equal difference sequence, which is the original three
It is known that three real numbers form an equal ratio sequence, and their product is 64. If the number in the middle is added with 1, it will form an equal difference sequence, which is the original three numbers. Find the three numbers

2,4,8 or 8,4,2
Let the original sequence be x / Q, x, X * Q
It can be known that X / Q * x * x * q = 64 (the middle term of equal proportion), so x = 4
The original sequence is 4 / Q, 4,4 * Q
Since the number in the middle is added with 1, it becomes an arithmetic sequence, then 5 * 2 = 4 / Q + 4q, and q = 2 or 1 / 2 is obtained
So the original sequence is 2,4,8 or 8,4,2

The first three of the four numbers are equal proportion series, and the last three are equal difference series. The first and last two terms are 21, and the middle two terms are 18. How to solve the four numbers?

Let the first number be a and the equal ratio be Q
a,aq,aq^2,2aq^2-aq
(1-q+2q^2)/(q+q^2)=21/18
Q = 2, or q = 3 / 5
So 3, 6, 12, 18
Or 75 / 4, 45 / 4, 27 / 4, 9 / 4

There are four numbers, of which the first three numbers form an equal difference sequence, and the last three numbers form an equal ratio sequence. The sum of the first and last two numbers is 21, and the sum of the middle two numbers is 18~

Let the four numbers be A-D, a, a + D, (a + D) ^ 2 / A in turn. According to the meaning of the title, they are: (A-D) + (a + D) ^ 2 / a = 21 （1）a+(a+d)=18………………………… (2) We get d = 18-2a from (2), and substitute (1) with a - (18-2a) + (a + 18-2a) ^ 2 / a = 21 = = = > 3a-18 + (18-a) ^ 2 / A-21 =

There are two numbers. The first three numbers form an equal ratio sequence, and the last three numbers form an equal difference sequence. The sum of the first and last two numbers is 21, and the sum of the middle two numbers is 18?

Let x, A-D, a, a + D, then x + A + D = 21x = 21-a-da-d + a = 18D = 2a-18, so x = 21-a-2a + 18 = 39-3a, the first three equal ratios (A-D) & sup2; = ax, so (18-a) & sup2; = a (39-3a) 324-36a + A & sup2; = 39a-3a & sup2; 4A & sup2; - 75A + 324 = 0 (4a-27) (A-12) = 0A = 27 / 4, a = 12a = 27 / 4, d =