Senior 1 math ,,, speed,, fast

One. ,0-1)=(1,-1)

b=4e1-e2=(4*1-0,-1)=(4,-1)a·b=1*4+(-1)*(1)=5

a+b=(1+4,-1-1)=(5,-2)a+b|= root number (5 + 2) = root number 29

2.Let the angle between A and B be

cosθ=(a·b)/(|a|*|b|) = 5 root number 34 two. 1.The waveform of the first pine nut falling:

The waveform of the second pine nut falling after y1 = 2sin(t 6)2s: y2 = 2sin((t-2) 6)y= 2sint 6 2sin(t-2) 6-2 root number 3sin((t-1) 6).

Root number 3sin (4 3) = 3m

1。Because a=e1-e2, b=4e1-e2 and e1=(1,0),e2(0,1).

So a=(1,-1) b=(4,-1).

So a*b=1*4+(-1)*(1)=5

So a+b=(5,-2).

So |a+b|= with the number 29

With the number 17

f(x)=-f(-x)

ax+b)/(1+x^2)=(ax-b)/(1+x^2)b=0f(x)=(ax)/(1+x^2)

1/2a)/(5/4)=2/5

a=1f(x)=x/(1+x^2)

If x2>x1

then f(x2)-f(x1)=x2 (1+x2 2)-x1 (1+x1 2).

x2+(x1^2)x2-x1-(x2^2)x1]/(1+x2^2)(1+x1^2)

x2-x1)(1-x1x2) (1+x2 2)(1+x1 2)>0 proposition holds. 3)f(t-1)=-f(1-t)f(t)-f(1-t)<0

f(t)t<1-t and spring beat-1 "guess t<1 and -1< T-1<10

m=｛-3，2｝

Since n is contained in m, then when y=-2 a=-3, we get a=2 3 or y=-2 a=2, and we get a=-1

That is, the value of the real number a is 2, 3, or -1

A intersection b = and a1 + a4 = 10, a1, a4 is the square of some two numbers and is an integer, a1a1 = 1, a4 = 9, a1 squared = 1, a4 squared = 81, a and b the sum of all elements is 124, that is, the sum of all the elements of a + the sum of all the elements of b - a intersection b the sum of all elements = a1 + a2 + a3 + a4 + a1 square + a2 + a3 square + a4 square - a1-a4

Get the square of A2 + A3 + A2 + the square of A3 = 42A4 is the square of A2 or A3, that is, A2 = 3, or A3 = 3 If A3 = 3, then A2 = 2, A2 + A3 + A2 square + A3 square = 18, round A2 = 3, substitute A2 + A3 + A2 square + A3 square = 42, A3 = 5A=, B=

In a, y=log(2)x, true number 0

Therefore x 0 m = (0,+.)

And when m=1b=[1,4].

a∩b=[1,4]

2.When b is contained in a, i.e. [m,m+3] a

There should be m 0 at this point

m∈(0,+∞

Because the angle c = 120 degrees.

So the angle a+b = 60 degrees.

Then tan(a+b) = (tana+tanb) (1-tanatanb) = 3

Substitution data is fine.

Solution: On (0, +infinity), take two independent variables x10, f(x2) > f(x1), that is, the function increases monotonically on (o, root number 2);

2) The root number 2 is a monotonic decrease in the function on [root number 2, positive infinity].

So when x = root number 2, the function has a maximum value.

f(root number 2) = 5 times root number 2 4

There is no minimum value for a function.

The maximum value can also be judged from another method.

x-root number 2) 2 0

x 2 + 2 times root number 2 * x

So f(x)=5x (x 2+2) 5x (2x root number 2) x = 5 times root number 2 4

When x = root sign 2, the equal sign holds, and the maximum value is 5 times the root sign 2 4