The equivalent condition of two vectors a and B being collinear

The equivalent condition of two vectors a and B being collinear

The equivalent condition for two vectors a and B to be collinear is
There are real numbers m and n such that Ma = NB holds
If a and B are plane vectors, and a = (x1, Y1), B = (X2, Y2)
Then the equivalent conditions of two vectors a and B being collinear are as follows:
x1·y2=x2·y1

If a and B are all nonzero vectors, under what conditions are vectors a + B and A-B collinear?
Let there be a real number λ such that
a+b=λ(a-b)=λa-λb
What is the discussion about 1 = λ?
I can't understand other analysis at this step. What's the situation

If we continue to sort out the above formula, we can get (λ - 1) a = (λ + 1) B. the reason why we discuss λ = 1 is that the zero vector is collinear with any vector. As long as the left side of the equation is a zero vector, then the formula holds. Similarly, we should also discuss λ = - 1

What is the necessary and sufficient condition for a vector to be perpendicular to B

The sum of the products of the vertical and horizontal coordinates equals zero

When a and B satisfy what conditions, | vector a + vector B | = | vector a | + | vector B | holds

Collinear, same direction

If a is a real number, then the absolute value of a is greater than 0. Is it right or wrong? Why?
P8.5.（3）

Wrong, 0 is a counterexample

It is known that Tan α and Tan β are the two real roots of the quadratic equation x * 2 + PX + 2 = 0 with respect to X
It is known that Tan α and Tan β are two real roots of the quadratic equation x ^ 2 + PX + 2 = 0 with respect to X. the following results are obtained: (1) Tan (α + β) = P (2) 3sin (α + β) + p * cos (α - β) = 0

1、tan（α+β）=（tanα+tanβ）÷(1-tanβtanα)；
Tan α, Tan β are the two real roots of the quadratic equation x * 2 + PX + 2 = 0 with respect to X,
Tan α + Tan β = - p; Tan α, Tan β = 2
tan（α+β）=（-p）÷（1-2）=p.
2. 3sin (α + β) + p * cos (α - β) = 3 (sin α cos β + cos α sin β) + P (COS α cos β + sin α sin β) = 0. By dividing both sides by cos α cos β, we can get 3sin (α + β) + p * cos (α - β) = 3 (Tan α + Tan β) + P (1 + Tan α Tan β);
As above, substituting Tan α + Tan β = - p; Tan α, Tan β = 2, which is equal to 3 (- P) + P (1 + 2) = 0, then it is proved

It is known that Tan θ and Tan (π / 4 - θ) are two roots of the quadratic equation x & sup2; - KX + 2k-5 = 0 with respect to x, where θ∈ (0, π / 2)
(1) Finding the value of K and the two roots of the equation

According to Weida's theorem, the two parts of the equation x & # 178; - KX + 2k-5 = 0 can be expressed as:
tanθ+tan(π/4-θ)=K
tanθ.tan(π/4-θ)=2K-5
And Tan (θ + π / 4 - θ) = {Tan θ + Tan (π / 4 - θ)} / {1-tan θ. Tan (π / 4 - θ)}
therefore
tan(θ+π/4-θ)=tanπ/4=1=K/(1-2K+5)
The solution is k = 2
The equation x & # 178; - KX + 2k-5 = 0 can be expressed as:
x²-2x-1=0
The solution is X1 = 1 + open radical (2); x2 = 1 - open radical (2)

Please explain that no matter what value a takes, the equation (A & sup2; - 4A + 5) x & sup2; - 6AX + 1 = 0 for X is a quadratic equation with one variable

a²-4a+5
=a²-4a+4+1
=(a-2)²+1
(a-2)²>=0
So (A-2) & sup2; + 1 > = 1
That is, no matter what the value of a is, the quadratic coefficient will not be equal to 0
So the quadratic equation of one variable

Given that the solution of equation 2 (x + 1) = 3 (x-1) is a + 2, find the solution of equation 2 [2 (x + 3) - 3 (x-a)] = 3A

From the solution of equation 2 (x + 1) = 3 (x-1), we get x = 5. Let a + 2 = 5, so a = 3. Then there is 2 [2 (x + 3) - 3 (x-3)] = 3 × 3, that is - 2x = - 21, ｜ x = 1012

If the equation (3 / 4) 2 = 3A + 2 / 5-a about X has negative roots, then the value range of a is?

Is it like this: (3 / 4) x = 3ax + 2 / (5-a)? Exponential function (3 / 4) ^ x = (3a + 2) / (5-a) has negative roots, which means x1, 3A + 2 > 5-a, 4A > 3, a > 3 / 4
Thank you!