【正确答案】含碳质量分数0.77/%＜X₃＜2.11/%铁碳合金室温下的组织为P+Fe₃C(珠光体+渗碳体)。
   w(P)=(6.69-X₃)/(6.69-0.77)×100/%
   w(Fe₃C)=(X₃-0.77)/(6.69-0.77)×100/%碳质量分数2.11/%＞X₄＞4.3/%铁碳合金在室温下组织为(P+Fe₃C)+L_d(P+C_m+Ld)
   第一步：w(A)=(4.33-X₄)/(4.33-2.11)×100/%
   w(L_d)=(X₄-2.11)/(4.33-2.11)×100/%
   第二步：w(P)=[(6.69-2.11)/(6.69-0.77)]×[(4.33-X₄)/(4.33-2.11)]×100/%
   w(Fe₃C)=(2.11-0.77)/(6.69-0.77)×100/%