Question

求解第四题。

Answer 1

解一元二次方程。

Answer 2

3*x=m,x㏒3=㏒m,m＞0,3*y=n,y㏒3=㏒n,n＞0,则m+n=w,m-n=6,
﹙x+y﹚㏒3=㏒m+㏒n=㏒mn,3㏒3=㏒mn,㏒27=㏒mn,mn=27
即有①m+n=w，②m-n=6，③mn=27,由②③解出n=3或m=9,n=-9﹙舍﹚，m=-3﹙舍﹚
∴m=9,n=3,即x㏒3=㏒9=2㏒3,x=2,
y㏒3=㏒3,y=1,w=3*2+3*1=12
∴x=2,y=1,w=12

Answer 3

x+y=3 (1)
3^x+3^y=w (2)
3^x -3^y =6 (3)
sub (1) into (3)
3^x -3^(3-x) =6
3^x -6 -27(3^(-x)) =0
[3^x -9(3^(-x)) ]. [3^x +3(3^(-x)) ] =0
3^x -9(3^(-x))= 0
3^x = 3^(2-x)
x=2-x
x= 1
from (1)
x+y=3
1+ y =3
y=2
from (2)
w=3^x+3^y
=3^1 + 3^2
=12