請教2題對數

1.
\(0<x<1<y<100<z\)且\(\displaystyle \cases{\displaystyle log_2 xyz=101 \cr \frac{1}{log_2 x}+\frac{1}{log_2 y}+\frac{1}{log_2 z}=\frac{1}{101}}\)，則\(xyz(x+y+z)-xy-yz-zx=\)？
答案：\(2^{202}-1\)

2.
點\(A(-4,3)\)，點\(P,Q\)各在\(f(x)=2^x,g(x)=log_2 x\)圖形上，且\(P,Q\)對稱於\(y=x\)直線，求\(\overline{AP}+\overline{AQ}\)的最小值？
答案：\(7\sqrt{2}\)
謝謝!!

第1題
\(\begin{align}
  & a={{\log }_{2}}x,b={{\log }_{2}}y,c={{\log }_{2}}z \\
& 0<x<1<y<100<z \\
& a<0,b>0,c>0 \\
&  \\
& a+b+c=101 \\
& \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{101}=\frac{1}{a+b+c} \\
& \frac{bc+ca+ab}{abc}=\frac{1}{a+b+c} \\
& \left( a+b+c \right)\left( bc+ca+ab \right)-abc=0 \\
& \left( a+b \right)\left( b+c \right)\left( c+a \right)=0 \\
& a+b=0\quad or\quad a+c=0 \\
& xy=1,z={{2}^{101}}\quad or\quad xz=1,y={{2}^{101}} \\
& ...... \\
\end{align}\)

第2題
\(\begin{align}
  & P\left( x,y \right),Q\left( y,x \right) \\
& \overline{AP}+\overline{AQ}=\sqrt{{{\left( x+4 \right)}^{2}}+{{\left( y-3 \right)}^{2}}}+\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y+4 \right)}^{2}}} \\
\end{align}\)
所求為\(\left( -4,3 \right)\)到\(\left( 3,-4 \right)\)之距離

能請教thepiano老師第一題後面的過程嗎?真的很不好意思!

\(\begin{align}
  & xy=1,z={{2}^{101}} \\
& xyz\left( x+y+z \right)-xy-yz-xz \\
& =z\left( x+y+z \right)-1-yz-xz \\
& =xz+yz+{{z}^{2}}-1-yz-xz \\
& ={{z}^{2}}-1 \\
& ={{2}^{202}}-1 \\
\end{align}\)

感謝thepiano老師!!