Xiao Gang may be ill. He can't come tomorrow

Xiao Gang may be ill. He can't come tomorrow

Inconsistency
Xiao Gang is ill. He can't come tomorrow

Xiao Gang recorded and listened to the speech at the meeting

Xiao Gang listened to and recorded the speech of the meeting

Xiao Gang recorded and listened to the speech of the meeting

Xiao Gang recorded and listened to the speech at the meeting
Cause: improper collocation. "Record" is not matched with "speech" and should be removed

Given nonzero vectors a and B, a = 1, (a-b) · (a + b) = 1 / 2
(1) B
(2) If a · B = 1 / 2, find the angle between vectors a and B

(a-b)·(a+b)=|a|²-|b|²=1/2
1 - | B | & sup2; = 1 / 2 | B | = root 2 / 2
2）ab=|a||b|cosβ
1 / 2 = root 2 / 2 cos β
Cos β = root 2 / 2
β=45°
The angle is 45 degrees

In triangle ABC, BD: DC = 3:4, AE: CE = 5:6, AF: FB =?

According to the same height triangle area ratio equal to the bottom ratio
△ABD:△ADC=BD:DC=3:4
△OBD:△ODC=BD:DC=3:4
The results show that △ oba: △ OCA = BD: DC = 3:4
Similarly, △ OBC: △ oba = Ce: EA = 6:5
The results show that △ OCA: △ OBC = (4 / 3): (6 / 5) = 10:9
That is AF: FB = 10:9

If f (x) = 1 / 3 x & # 179; + | a | X & # 178; + 2A * BX + 1 has extremum on R, then the value range of F (x) = 1 / 3 x & # 179; + | a | X & # 178; + 2A * BX + 1 has extremum on R
Seek detailed process

f(x)=1/3 x³+|a|x²+2a*bx+1
Its derivative function is: F '(x) = x & # 178; + 2|a|x + 2A * B
=x²+2|a|x+2|a|*|b|*cos
∵ f (x) has an extreme value on R, that is, f '(x) = 0 has a real root,
The ⊿ = (2 | a |) & 178; - 4 * 1 * 2 | a | * | B | * cos ≥ 0 of the equation x & # 178; + 2 | a | x + 2 | a | * | * B | * cos = 0
And | a | = √ 3 | B |,; ⊿ = 12 | B | & # 178; - 8 √ 3 | B | & # 178; Cos = 4 √ 3 | B | & # 178; (√ 3-2cos) ≥ 0
It is known that a and B are nonzero vectors, and | B | & # 178; > 0 is constant, and ⊿ = 4 √ 3 | B | & # 178; (√ 3-2cos) ≥ 0 is constant,
That is √ 3-2cos ≥ 0, ｜ cos ≤ √ 3 / 2,
The angle θ formed by ∵ plane vector is: θ ∈ [0, Π], ∵ θ ∈ [Π / 6, Π]
Conclusion: - 1 ≤ cos ≤ √ 3 / 2 trigonometric function value range
The value range of the angle between θ∈ [Π / 6, Π] vectors
Hope to help readers to solve their doubts!

On proving that vectors are coplanar
I don't seem to have taken all the notes that the teacher wrote
Vector AC = λ 1, vector AB + λ 2, vector ad has no specific requirements for λ 1, λ 2
Which points can be proved coplanar by this condition? What's the relationship between point a and BCD?

Vector AC = λ 1 vector AB + λ 2 vector ad
There is no specific requirement for λ 1 λ 2
This condition proves that point a is coplanar with B, C and D
A point and B, C, D relation? I don't know what you say
But if there is
Vector AC = λ 1 vector AB + λ 2 vector AD and λ 1 + λ 2 = 1
Then B, C and D are collinear

As shown in Figure 12, in △ ABC, ∠ ACB = 90 °, both △ ADC and △ BEC are equilateral triangles, and the extended DC intersects at point F. please explain:

Conclusion is not clear, can only guess to write
If ∠ ACB = 90 ° and ∠ ACD = 60 °, then: ∠ BCF = 180 ° - ∠ ACD - ∠ ACB = 30 °;
And ∠ CBE = 60 °
So ∠ CFB = 90 ° and CF ⊥ be is obtained;
And CB = CE, so BF = EF

The following statement is wrong. A, usually use x, y, Z to represent the unknowns. B, set letters to represent the unknowns. C, solving the equation is to find the value of the unknowns that make the left and right sides of the equal sign in the equation equal. D, the solution of an equation is x = 1, "x = 1" is no longer a unary linear equation. That is wrong, why?

D
x=1
An unknown number, and the unknown number is once
So he's an equation of first degree with one variable

If vector a = (3, - 4), then the absolute value of vector a=

Vector a = (3, - 4)
Then | a | = √ (3 ^ 2 + (- 4) ^ 2) = 5
This is not the absolute value, but their modulus, that is, the length!
Do not understand can ask, thank you!