Freshman Math! Hurry, hurry! Process!

Solution: (1) Because 3 x=4 y=6 z, then xlg3=ylg4=zlg6, because 2x=py

So p=2x y=2lg4 lg3

2log(3)4.

2) Let xlg3=ylg4=zlg6=k, then 1 x=lg3 k, 1 y=lg4 k, 1 z=lg6 k

So 1 Z-1 X=(Lg6-Lg3) KlG2 K,1 2Y=(1 Y) 2

lg4/k)/2

2lg2/2k

LG2 K, so 1 Z-1 X=1 2Y

I'll do a question.

17.(1)f(x)=x+9/(x-3)，x>3,=(x-3)+9/(x-3)+3

2 9+3=9, when x-3=9 (x-3), i.e. x=6, take the equal sign, so the minimum value of f(x) is 9

2) f(x) t (t+1)+7 is constant, which is equivalent to 2-t (t+1) 0, which is equivalent to (t+2) (t+1) 0, so t>-1 or t -2 is what is sought.

1、f(x)min=9; (x-3)+9/(x-3)+3>=3+2[(x-3)*9/(x-3)]^1/2)=9;

2. Substituting f(x)min=9 into the inequality equation; We get t>=-2 and t≠-1.

Solution:1Take the midpoint E of AB and connect EF.

Since E and F are the midpoints of PB and AB respectively, then in the triangle PAB, PA is parallel to EF, and in the quadrilateral ADEF: the triangle ADE is similar to the triangle FED both ADE= FED, AED= FDE, DAE= EFD and the bottom ABCD is square, then DAE=90°, so EFD=90°

then EF DF

Because of EF AP

So DF AP

2.Exist.

Do the midpoint i on the pa, then since pd=pa, the angle pada is a right angle, so di is perpendicular to pa, if is the triangular pab median, so if is parallel to ab, ab is perpendicular to ad and pd, so, ab is perpendicular to the surface pda, so ab is perpendicular to pa, so if perpendicular to pa. Since id is perpendicular to pa, pa is perpendicular to face ifd, so pa is perpendicular to df

Let 3 x=4 y=6 z=k

1) x=log3(k),y=log4(k)p=2x y=2log3(k) log4(k)=2log3(k)·logk(4)=2log3[k logk(4)]=2log3(4);

2)1 Trace Z-1 x=logk(6)-logk(3)=logk(6 3)=logk(2)=logk[4 (1 2)]=logk(4)] 2=1 [2log4(k)]=1 (2y).