If | a | = 1, | B | = 2, C = a + B, and C ⊥ a, then the angle between a and B is Hope there are detailed steps

If | a | = 1, | B | = 2, C = a + B, and C ⊥ a, then the angle between a and B is Hope there are detailed steps

Let the angle be α, then cos α = (vector a × b) / (/ A / * / B /) / denotes the absolute value
/a/*/b/=1*2=2
Let a = (Q, w), B = (E, R), C = (T, y)
Then q ^ 2 + W ^ 2 = 1
e^2+r^2=4
t=q+e,y=w+r
qt+wy=0
The solution is Q (Q + e) + W (W + e) = 1 + QE + WR = 0
Because the vector a × B = QE + WR = - 1
So cos α = - 1 / 2
So α = 120 degrees

Vector a × B × C = a × (B × C) = (a × b) × C = Mody?
a×（b×c）=（ac)b-（ab）c
a×b×c=?

a×（b×c）=（ac)b-（ab）c
（ac)b-（ab）c
=ABC ABC (direct de bracket)
=0
So a × (B × C) = ABC = 0
This formula is 0-0 = 0

A and B solve the equations ax + 5Y = 15 ①, 4x + by = - 2 ② together. Because a misread the letter A, we get the equations ① x = - 3, y = - 1
If B misreads the letter B, the solution of the original equation is x = 5, y = 4

X = - 3, y = - 1 is the solution of the equation 4x + by = - 2
∴-12-b=-2
b=-10
X = 5, y = 4 are the solutions of the equation AX + 5Y = 15
∴5a+20=15
∴a=-1
The original equations are
-x+5y=15
4x-10y=-2
∴x=14
y=5.8
∴x-y=8.2

A and B solve the equations ax + 5Y = 15, 4x + by = - 2 together. Because a misreads the letter A, we get the equations x = - 3, y = - 1
The solution of the equations is x = 5, y = 4, try to calculate the value of A2008 power + B

Because a misread equation 1, equation 2 is correct, that is, the solution x = - 3, y = 1 is suitable for equation 2;
Substituting x = - 3, y = 1 into 2, we get 4 * (- 3) + b * 1 = - 2, and get b = 10
Similarly, B misread equation 2, that is, equation 1 is correct; substituting x = 5, y = 4 into equation 1, a * 5 + 5 * 4 = 15, a = - 1 is solved
From the known
Substituting x = - 3, y = 1 into 2, we get 4 * (- 3) + b * 1 = - 2, and get b = 10
Substituting x = 5, y = 4 into equation 1, a * 5 + 5 * 4 = 15, a = - 1 is obtained
So A2008 power + B = 1-4 = - 9

A and B solve the equations ax + 5Y = 15, 4x + by = - 2 together. Because a misreads the letter A, we get the equations x = - 3, y = - 1. B looks at the letter B, we get the equations
The solution of A-B is x = 5, y = 4

X = - 3Y = - 1 is the solution of the equations ax + 5Y = 15,4x + by = - 2, which can be substituted into the equations - 3a-5 = 15, - 12-B = - 2; a = - 20 / 3, (wrong) B = - 10, respectively
X = 5Y = 4 is the solution of the equations ax + 5Y = 15,4x + by = - 2, which can be substituted into the equations 5A + 20 = 15,20 + 4B = 4; a = - 1, B = - 4 respectively (wrong)
a-b=-1-(-10)=9

Let K be a real number, and x1x2 be the two roots of the equation x ^ 2 + kx-1. If (| x1 | - x2) (| x2 | - x1) ≥ 1, then the value range of K

x1*x2=-1

We know the quadratic equation x ^ + kx-1 = 0 with respect to X. let two X 1, x 2 of this equation satisfy x 1 + x 2 = x 1, x 2, and find K

According to Weida's theorem, X1 + x2 = - K, x1x2 = - 1,
And X1 + x2 = x1x2,
So k = 1

Two different signs of x 1 x 2 of quadratic equation with one variable x 2-k x + 5 (K-5) = 0 and satisfying 2x1 + x 2 = 7 to find K

x1+x2=k
So 2x1 + x2 = X1 + k = 7
x1=7-k
So 7-K is the root of the equation
Substituting
49-14k+k²-7k+k²+5k-25=0
k²-8k+12=0
(k-2)(k-6)=0
Otherwise, x1x2 = 5 (K-5)

Given that x1x2 is the two real roots of the equation x ^ 2-kx + 1 / 4K (K + 4) = 0 about X, when k takes what value, (x1-2) (x2-2) = 37 / 4

Is that 37 / 4? If so, K1 equals 7, K2 equals - 3

1. The univariate quadratic equation x & sup2; + kx-1 = 0 about X is known. Let two of the equations be X1 and X2 respectively, and satisfy X1 + x2 = x1x2, and find the value of K

The results of Weida's theorem are as follows
x1+x2=-k
x1x2=-1
So, - k = - 1
k=1