九师联盟 2024届江西高一下学期开学考数学.考卷答案试卷答案,我们目前收集并整理关于九师联盟 2024届江西高一下学期开学考数学.考卷答案得系列试题及其答案，更多试题答案请关注微信公众号：考不凡/直接访问www.kaobufan.com(考不凡)
九师联盟 2024届江西高一下学期开学考数学.考卷答案试卷答案
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MyTeacherMr.MooreThere'sateacherMr.Moore.Whoislovelyandthirty-four.Alwaysencouragingustotry.Heleadsustoaworldof“why”.Wealladmirehimmoreandmore.Theauthorofthispoemisunknown,butitremindsmeofthecountlessteacherswhohavetaughtme.WheneverIreadthispoem,Icouldn'thelpthinkingofmymathteacher,Mrs.Davies.Ifirmlybelievethatshewassuchateacherdescribedinthepoem.Inhighschool,Mrs.Davieswasmymathteacher,whotookherjobseriouslyandexpec-tedherstudentstoputforththesamedegreeofcommitment.Thoughshewas30,shelookedveryyoungforherage.Andshewasoneofthemostpopularteachersinourschool.Comparedwithotherteachers,Mrs.Daviespaidmoreattentiontoherwayofteaching.Shetriedvariouswaystomakeherclasseslivelyandinteresting.Strictbutfair,sheheldourattentionwithhertime-provenapproach.Oneofherclasseswasforstudentswhoweremoreadvancedacademically.Ichosetotakeitandsoonrealizedthatthecoursewasabitovermyhead.Istruggledwiththecomplexproblemsthatothersseemedtolearnwithoutmucheffort.OneFridayweweretestedonourabilitytouseatheorem(thatnooneintheclassthoroughlyunderstood.Theteacherwouldgradeourpapersbasedonourabilitytoprogressthroughtheproblem.Iwasabsolutelylost.Finally,Ididn'tusetherequiredtheorem.Instead,Idecidedtousemorefamiliartheoremstoarriveattheanswer.IknewIhadfailedbecauseIhadn'tdonetherequiredassignment.Ibecameresignedtomyfate.注意：1.续写词数应为150左右；2.请按如下格式在答题卡的相应位置作答
OurtestpaperswerehandedbackthefollowingMonday.ThoughIdidn'tusetherequiredtheorem,Mrs.Daviesseemedgenuinelyproudofme.【高三英语第10页（共10页）】·23-93C·
分析对x${\;}^{\frac{2}{3}}$+y${\;}^{\frac{2}{3}}$=a${\;}^{\frac{2}{3}}$两边对x求导，可得$\frac{2}{3}$${x}^{-\frac{1}{3}}$+$\frac{2}{3}$${y}^{-\frac{1}{3}}$•y′=0，代入切点的坐标，可得斜率，再由点斜式方程，可得切线的方程．
解答解：对x${\;}^{\frac{2}{3}}$+y${\;}^{\frac{2}{3}}$=a${\;}^{\frac{2}{3}}$两边对x求导，可得
$\frac{2}{3}$${x}^{-\frac{1}{3}}$+$\frac{2}{3}$${y}^{-\frac{1}{3}}$•y′=0，
即有y′=-$\frac{{x}^{-\frac{1}{3}}}{{y}^{-\frac{1}{3}}}$，
可得在点（$\frac{\sqrt{2}}{4}$a，$\frac{\sqrt{2}}{4}$a）处的切线斜率为k=-1，
则在点（$\frac{\sqrt{2}}{4}$a，$\frac{\sqrt{2}}{4}$a）处的切线方程为y-$\frac{\sqrt{2}}{4}$a=-（x-$\frac{\sqrt{2}}{4}$a），
即为x+y-$\frac{\sqrt{2}}{2}$a=0．
点评本题考查导数的运用：求切线的方程，考查直线方程的求法，两边同时对x求导是解题的关键．
