Two high school math problems, you can add points to the logarithmic function

1.Let the function f(t)=(log2 3) t-(log5 3) tlog2 3>log2 2=1, 0=log5 1=(log2 3) -y -(log5 3) -y

Know f(x)>=f(-y).

then x>=-y

x+y>=0

Select (b).

2.The two roots of (lgx) 2-lgx 2-2=0 are A and B, then (LGA) 2-2*LGA-2=0, LGB) 2-2*LGB-2=0

Then (LGA) 2=2LGA+2

lgb)^2=2lgb+2

From Vedic theorem, LGA+LGB=2 can be obtained

lga*lgb=-2

then loga b + logb a

LGB LGA)+(LGA LGB) [(LGB) 2+(LGA) 2] (LGA*LGB)[2LGA+2+2LGB+2] (-2).

2(lga+lgb)+4]/(-2)

1.Option B, Reason, log2(3)] x-[log5(3)] x [log2(3)] y)-[log5(3) (y),log2(3)] x-[log5(3)] x [(log5(3)) y*(log2(3)) y], log5(3)) y>0, (log2(3)) y>0, yes.

log5(3)) y*(log2(3)) y]*{log2(3)] x-[log5(3)] x [log5(3)] y-log2(3)] y,log2(3)] x+y)*[log5(3)] y-[log5(3)] x+y)*[log2(3)] y [log5(3)] y-log2(3)] y, only if [log2(3)] x+y)*[log5(3)] y [ log5(3)] y,log2(3)] x+y) 1=[log2(3)] 0, i.e., x+y 0,2(lgx) 2-lgx 2-2=0, lgx-1) 2=3, lgx= 3+1, or lgx=- 3+1

x1=10^(√3+1),x2=10^(1-√3).

ream, a=x1, b=x2

then loga b + logb a = (1 - 3) (1 + 3) + (1 + 3) (1 - 3) = 8 -2 = -4

NND and what I learned in school have all been returned.

Solution: According to the known straight line oc: y=x straight line ab: y=-x+6024When

It's easy to divide it into three sections.

1.(1) Prove: f(x+2)=-f(x+1)=f(x) 2 is a period of f(x).

2) Let x be in the interval [2k-1,2k+1],k z, then x-2k [-1,1], and know that 2 is a period of f(x), so 2k is also its period (which can be proved by itself), and when x [-1,1], f(x)=x 2, so f(x)=f(x-2k)=(x-2k) 2 (x [2k-1,2k+1],k z).

2.(1) Find the domain of f(x) first, x+2 is not equal to 0 and (x-3) (x+2)>=0 obtains: x>=3 or x<-2

Therefore a=(2) If a=0, then the domain b is defined as x>b 2, and it is clear that b cannot include all a, so a is not equal to 0.

From (2x-b)(ax+1)>0 and b>0, we can see that the solution of the inequality is x>b 2 or x<-1 a, and b 2>-1 a is constant, and a>0

And a belongs to b, so b 2 < = 3, -1 a> = -2 gets: b< = 6, a> = 1 2 or a<0 (rounded).

So a>=1 2,0 high school math has been lost for a long time, I hope it will help you, if you are wrong, please forgive me!

1.(1)f(x+2)=-f(x+1)=f(x) Of course, 2 is the cycle.

2) 2 is the period, then 2k is also the period. When x [-1,1], f(x) = x 2, so f(x) = f(x-2k) = (x-2k) 2 (x [2k-1,2k+1],k z).

2.(1) The definition field x+2 is not equal to 0, and (x+7) (x+2) is less than or equal to 2, then x is less than -2, or, x is larger.

is equal to 32) g(x) defines the domain b uncertain, and when a is 0 and does not hold, then a is not 0.

From (2x-b)(ax+1)>0 and b>0, the solution of the inequality is x>b 2 or x<-1 a, and b 2>-1

A is constant, then a is greater than 0

A belongs to b, then 2/2 b is less than or equal to 3, and 1/1 of -a is greater than or equal to -2

i.e. b<=6, a>=1 2 or a<0

a<0 is not valid.

a>=1/2,0

Question 1 For any x, y r has f(x+y)+f(x-y)=2f(x)f(y), and f(0)≠0

Let x=y=0 give f(0)=1

f(x+y)+f(x-y)=2f(x)f(y), so that x=0 has f(y)+f(-y)=2f(0)f(y), f(y)=f(-y).

f(x) is an even function.

The second question is a=b f(a b)=f(a)-f(b)∴f(1)=0∵f(4)=1 ∴f（16／4)=f(16)-f(4).

f(16)=2 f(x+6)-f(1 x) 2 =f(16)x+6>0,1 x>0 x(x+6)>16 gives x>2

Let 4-x=0, x=4, then f(4-x) over (4,1) and the inverse function image over (1,4).

The power of 3 5 power, first cubic and then 5 power, ordinary calculators can't calculate, computer calculators calculate yes.