%%% fig.tex 2016-4-25 8:55 %%% fig.sce 2016-4-25 8:55 {\unitlength=8mm% \begin{picture}% ( 10.22000, 3.87000)( -0.61000, 2.03000)% \special{pn 8}% % \settowidth{\Width}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}\setlength{\Width}{0\Width}% \settoheight{\Height}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}\settodepth{\Depth}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}% \put(0.0500,5.0000){\hspace*{\Width}\raisebox{\Height}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}}% % % \settowidth{\Width}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Width}{0\Width}% \settoheight{\Height}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\settodepth{\Depth}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}% \put(0.0500,3.0000){\hspace*{\Width}\raisebox{\Height}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}}% % % \end{picture}}%