当前位置: X题卡 > 所有题目 > 题目详情

单室模型静脉滴注给药过程中体内血药浓度随时间变化的关系式

查看本题答案

你可能感兴趣的试题

logX=(-K/2.303)t+logX0  logC=(-K/2.303)t+logC0  logC′=(-K/2.303)t′+log(K0/VK)  logC′=(-K/2.303)t′+log[K0(1-e-KT)/VK]  C=K0(1-e-KT)/KV  
C=C0(1-e-kt)/Vk  logC’=(-k/2.303)t’+log(k0/Vk)  logC’=(一k/2.303)t’+log(k0(1-e-kt)/Vk  logC=(一k/2.303)t+logC0  logX=(一k/2.303)t+logX0  
logX=(-K/2.303)t+logX0   10gC=(-K/2.303)t+logC0   logC′=(-K/2.303)t′+log(K0/VK)   10gC′=(-K/2.303)t′+log[K0(1-e-KT)/VK]   C=K0(1-e-Kt)/KV  
logX=(-K/2.303)t+logX0  logC=(-K/2.303)t+logC0  logC′=(-K/2.303)t′+log(K0/VK)  logC′=(-K/2.303)t′+log[K0(1-e-KT)/VK]  C=K0(1-e-KT)/KV  
C=C0(1-e-kt)/Vk  logC’=(-k/2.303)t’+log(k0/Vk)  logC’=(一k/2.303)t’+log(k0(1-e-kt)/Vk  logC=(一k/2.303)t+logC0  logX=(一k/2.303)t+logX0  
logX=(-K/2.303)t+logX0  10gC=(-K/2.303)t+logC0  logC′=(-K/2.303)t′+log(K0/VK)  10gC′=(-K/2.303)t′+log[K0(1-e-KT)/VK]  C=K0(1-e-Kt)/KV  
logX=(-K/2.303)t+logX0   10gC=(-K/2.303)t+logC0   logC′=(-K/2.303)t′+log(K0/VK)   10gC′=(-K/2.303)t′+log[K0(1-e-KT)/VK]   C=K0(1-e-Kt)/KV  
logX=(-K/2.303)t+logX0  10gC=(-K/2.303)t+logC0  logC′=(-K/2.303)t′+log(K0/VK)  10gC′=(-K/2.303)t′+log[K0(1-e-KT)/VK]  C=K0(1-e-Kt)/KV  
单室模型  双室模型  静脉注射给药  等剂量、等间隔  血管内给药  静脉滴注给药  
C=K0(1-e-kt)/VK  logC’=(-K/2.303)t’+log(K0/VK)  logC’=(-K/2.303)t’+log(K0(1-e-kt)/VK)  logC=(-K/2.303)t+logC0  logX=(-K/2.303)t+logX0  
C=k(1-e)/Vk  logC′=(-k/2.303)t′+log(k/vk)  logC′=(-k/2.303)t′+log[k(1-e)/Vk)]  logC=(-k/2.303)t+logC  logX=(一k/2.303)t+logX  
logX=(-K/2.303)t+logX0  logC=(-K/2.303)t+logC0  logC′=(-K/2.303)t′+log(K0/VK)  logC′=(-K/2.303)t′+log[K0(1-e-KT)/VK]  C=K0(1-e-KT)/KV  
logX=(-K/2.303)t+logX0   10gC=(-K/2.303)t+logC0   logC′=(-K/2.303)t′+log(K0/VK)   10gC′=(-K/2.303)t′+log[K0(1-e-KT)/VK]   C=K0(1-e-Kt)/KV  
C=K0(1-e-kt)/VK  logC’=(-K/2.303)t’+log(K0/VK)  logC’=(-K/2.303)t’+log(K0(1-e-kt)/VK)  logC=(-K/2.303)t+logC0  logX=(-K/2.303)t+logX0  
C=K0(1-e-kt)/VK  logC’=(-K/2.303)t’+log(K0/VK)  logC’=(-K/2.303)t’+log(K0(1-e-KT)/VK)  logC=(-K/2.303)t+logC0  logX=(-K/2.303)t+logX0  
logX=(-K/2.303)t+logX0   10gC=(-K/2.303)t+logC0   logC′=(-K/2.303)t′+log(K0/VK)   10gC′=(-K/2.303)t′+log[K0(1-e-Kt)/VK]   C=K0(1-e-Kt)/KV  
C=K0(1-e-kt)/VK  logC’=(-K/2.303)t’+log(K0/VK)  logC’=(-K/2.303)t’+log(K0(1-e-kt)/VK)  logC=(-K/2.303)t+logC0  logX=(-K/2.303)t+logX0