Mathematics ,,, in the first year of high school, and seek detailed analysis

,x∈[1,5]

So f(2x-3) = 2(2x-3) + 1 = 4x-5

This is because f(x) is defined in the domain [1,5].

So 1 2x-3 5

4≤2x≤8

The domain of 2 x 4 i.e. f(2x-3) is [

So the function f(2x-3) = 4x-5, x [2,4],2Because a-(a-1)=1

So [a-(a-1)] 2=1

i.e. a 2-1 + (a -2) -1

a^2+(a^-2)=2

So the original formula = (a 3 + a -3) [a 2 + (a -2)-3] (a 4-a -4).

a+a^-1)(a^2+a^-2-1)[a^2+(a^-2)-3]/(a^2+a^-2)(a^2-a^-2)

a+a^-1)(a^2+a^-2-1)[a^2+(a^-2)-3]/(a^2+a^-2)(a+a^-1)(a-a^-1)

a^2+a^-2-1)[a^2+(a^-2)-3]/(a^2+a^-2)(a-a^-1)

3.Because of the two solid roots, it satisfies 0 1 2

So f(0)f(1)<0

f(1)f(2)<0②

Solution 3t+3-7t+4<0

3-4t+4<0

t>7/4

Solution (7-4t)(12t+6-14t+4)<0

7-4t)(-2t+10)<0

4t-7)(2t-10)<0

7 4 so the range of t is (7 4, 5).

The obtuse angle of your formula is less than -1, and everything else is correct.

Then the acute angle is greater than 0 and less than 1

When equal to 1 is on a straight line, equal to 0 is a right angle.

The title may be the first n terms of the query.

1.First, find an=sn-s(n-1)=104-3n2Then find the piecewise function tn

When n<=34, tn=104n-3n(n+1) 2When n>34, tn=5287+3(n+35)(n-34) 2-104n

Because 2< 6< 3+1

sina c 2=( 3+1) 2=4+2 3<10=2 2+( 6) 2=a 2+b 2

ABC is an acute triangle, thus a< b< c should therefore be filled in: c

Have you ever studied the cosine formula? cosc=(a +b -c) 2ab, where c is the longest side, and the corresponding maximum angle is the angle c, substituting the formula, according to the calculator to know c=75°

Hello landlord.

First of all, f(x) defines the domain as [0,2], which means that the value of x is [0,2] Can this step be understood?

Then, since the above equation holds, g(x) is regarded as f(2x) and the other, then f(2x) is regarded as a whole in parentheses, and the value of 2x is [0,2], then the value of x is [0,1].

g(x) is a fractional function, so the denominator is not zero, so x is not equal to 1 and the definition domain of gx is [0,1).

Got it? I don't know how to ask, this is to find the definition domain of the composite function, in fact, it is not difficult to see the parentheses as a whole, it is not difficult, the main thing is to understand how the function f(x) is defined

[0,1) From the question, it can be seen that the domain [0,2] then 2x belongs to [0,2] x belongs to [0,1] and x-1 is not equal to 0

Draw the image of the function y, and since it is an absolute value, you can fold the image below the x-axis to the top of the x-axis.

Obviously, if the range is 0,1, then the domain m,n must be a subset of 1 a,a, and 1 a m 1 is defined because n m is a minimum of 5 6.

Obviously, m 1 a, n 1 or m 1, n a can get the minimum value of n m, and the solution is a 6 or 11 6

Select by=lnx-1 x as the single increment function.

f(1)=0-1=-1<0 f(e)=1-1/e>0

So the zero point belongs to (

Because c is a set of numbers, and b and d are sets of points, it doesn't matter.

After the simplification of the formula in b, it is (x-2y)(x+y)=0, so it can be seen that when x-2y or x+y is equal to zero, the formula b is true, so b can be rewritten so that the set d is a true subset of b.

Answer: Two straight lines are parallel, and the coefficients before x and y are proportional.

That is, 1:2m=(1+m):4

4=2m(1+m)

4=2m+2m²

m²+m-2=0

m = 1 or m = -2

1) When m=1, the straight lines are x+2y=0 and 2x+4y+16=0, and the two lines are parallel.

2) When m=-2, the straight lines are x-y=4 and -4x+4y+16=0, and the two straight lines coincide.

m=1 when the two lines are parallel.

Pick B

Parallel is illustrated.

1. The slopes of the two straight lines are equal.

k1=-1 (m+1), k2=-(1 2)*mlk1=k2 then find m=-2 or 1

2. Both lines are parallel to the y-axis (neither slope exists at this time).

The expression of such a line should be x=a (a constant), and the straight line 2 obviously does not satisfy the sum m=-2 or 1

The answer is a