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學(xué)習(xí)快評丨qibao666888　　官??(guan)方??(fang)數(shù)??(shu)據(jù)??(ju)顯??(xian)示??(shi)??，2024年??(nian)MXGP中??(zhong)國??(guo)站??(zhan)比??(bi)賽??(sai)為??(wei)上??(shang)海??(hai)帶??(dai)來??(lai)直??(zhi)接??(jie)經(jīng)??(jing)濟??(ji)效??(xiao)益??(yi)達??(da)1.14億??(yi)元??(yuan)??，間??(jian)接??(jie)經(jīng)??(jing)濟??(ji)效??(xiao)益??(yi)高??(gao)達??(da)2.68億??(yi)元??(yuan)??，獲??(huo)評??(ping)“2024年??(nian)上??(shang)海??(hai)體??(ti)育??(yu)消??(xiao)費??(fei)節(jié)??(jie)典??(dian)型??(xing)案??(an)例?(li)”?。??據(jù)??(ju)介??(jie)紹??(shao)??，2025MXGP與??(yu)F1聯(lián)??(lian)動??(dong)?。這??(zhe)兩??(liang)項??(xiang)世??(shi)界??(jie)頂??(ding)級??(ji)機??(ji)車??(che)賽??(sai)事??(shi)的??(de)聯(lián)??(lian)動??(dong)??，不??(bu)僅??(jin)可??(ke)為??(wei)車??(che)迷??(mi)帶??(dai)來??(lai)雙??(shuang)重??(zhong)盛??(sheng)宴??(yan)??，也??(ye)將??(jiang)進??(jin)一??(yi)步??(bu)推?(tui)動??(dong)上??(shang)海?(hai)建??(jian)設(shè)??(she)國??(guo)際?(ji)一??(yi)流??(liu)體??(ti)育??(yu)賽??(sai)事??(shi)之??(zhi)都??(dou)的??(de)進??(jin)程?(cheng)??，并??(bing)為??(wei)上??(shang)海??(hai)帶??(dai)來??(lai)顯??(xian)著?(zhu)的??(de)經(jīng)??(jing)濟??(ji)社??(she)會?(hui)影??(ying)響??(xiang)?。　　2025年??(nian)將?(jiang)是??(shi)這??(zhe)項?(xiang)全??(quan)球??(qiu)矚??(zhu)目??(mu)的?(de)摩??(mo)托??(tuo)車??(che)賽??(sai)事??(shi)第??(di)三??(san)次??(ci)來?(lai)到??(dao)中??(zhong)國?(guo)??，并??(bing)將??(jiang)在??(zai)未??(wei)來??(lai)持??(chi)續(xù)?(xu)舉??(ju)辦??(ban)??。??2019年??(nian)和??(he)2024年??(nian)??，MXGP中??(zhong)國??(guo)站??(zhan)的??(de)成??(cheng)功??(gong)舉??(ju)辦??(ban)??，獲??(huo)得??(de)了??(le)國??(guo)際??(ji)摩??(mo)托??(tuo)車??(che)聯(lián)??(lian)合??(he)會??(hui)(FIM)的?(de)高?(gao)度?(du)贊??(zan)譽??(yu)??。據(jù)??(ju)悉??(xi)??，MXGP每??(mei)年??(nian)在??(zai)全??(quan)球??(qiu)舉??(ju)辦??(ban)20個??(ge)分?(fen)站??(zhan)賽??(sai)??，最??(zui)終??(zhong)根??(gen)據(jù)??(ju)積??(ji)分??(fen)決??(jue)出??(chu)年??(nian)度??(du)車??(che)手??(shou)和??(he)車??(che)隊??(dui)總??(zong)冠??(guan)軍??(jun)??。將??(jiang)在??(zai)上??(shang)海??(hai)舉??(ju)行??(xing)的??(de)中??(zhong)國??(guo)站??(zhan)比??(bi)賽??(sai)作?(zuo)為祝(wei)MXGP年??(nian)度??(du)賽??(sai)事??(shi)的??(de)倒??(dao)數(shù)??(shu)第??(di)二??(er)站??(zhan)??。。