解：?$(1)(5\ \mathrm {m^2}-4m+2)-(4\ \mathrm {m^2}-4m-7)=5\ \mathrm {m^2}-4m+2-4\ \mathrm {m^2}+4m+7=\mathrm {m^2}+9$?
∵不論?$m $?為何值，?$\mathrm {m^2}+9＞0$?恒成立
∴?$5\ \mathrm {m^2}-4m+2＞4\ \mathrm {m^2}-4m-7$?
?$(2)$?∵?$A=5\ \mathrm {m^2}-4(\frac {7}{4}m -\frac {1}{2})$?，?$B=7(\mathrm {m^2}-m)+3$?
∴?$A-B=[5\ \mathrm {m^2}-4(\frac {7}{4}m -\frac {1}{2}) -[7(m2-m)+3]$?
?$=5\ \mathrm {m^2}-4(\frac {7}{4}m -\frac {1}{2})-7(\mathrm {m^2}-m)-3$?
?$=5\ \mathrm {m^2}-7m+2-7\ \mathrm {m^2}+7m-3$?
?$=-2\ \mathrm {m^2}-1$?
∵不論?$m $?為何值，?$-2\ \mathrm {m^2}-1＜0$?恒成立
∴?$A-B＜0$?，即?$A＜B$?
?$(3)(3a+2b)-(2a+3b)=3a+2b-2a-3b=a-b$?
當(dāng)?$a＞b$?時，?$a-b＞0$?，此時?$3a+2b＞2a+3b$?
?$ $?當(dāng)?$a=b$?時，?$a-b=0$?，此時?$3a+2b=2a+3b$?
當(dāng)?$a＜b$?時，?$a-b＜0$?，此時?$3a+2b＜2a+3b$?
