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請教2題對數

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

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回復 3# thankyou 的帖子

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

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