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8.C【解析】当木板与水平面夹角为53°时,两物块剛好滑动,A有沿B上表面向上运动的趋势,A、B间的静摩擦力f1=FN=mgos53°①,根据平衡条件可知绳对A的拉力T=mgsIn53°+f1②,此时B有沿木板面向下的运动趋势,B与木板间的静摩擦力f2=N=·6mgcOs53°③,以B为研究对象,根据平衡条件可知,5mgin53=T+f+/2④,联立①②③④解得p=2,选项C正确
2.(1)证明:∵an+1=Sn+1-Sn,S2=an+1-Sn+1S)-ASSutI(s(2)存在∵Sn+1=25+λ∴Sn=25-1+A(n≥2),相减得an+1=2an(n≥2),∴{an}从第二项起成等比数列,∵S2=2S1+A,即a2+a1=2a1+入(入+1)2若使{an}是等比数列,则a1a3=a2,∴2(A+1)=(A+1)2λ=1,经检验,符合题意故存在实数λ,使得数列{an}为等比数列,λ的值为1
4.【答案】B【解析]知数列a.=2n+1,其前m项的和S.=2×1+1+2n+Dm=m(n+2),则s.m(n+2)2mm+2)所S,S,)==(1+故选B324nn+222n+1n+2
15.【答案】2【解析】由S,S,S4成等差数列,得2S,=S3+S,设等比数列{an}+40,刚1所以2(-7得0,又因为的公比q=1,则S,=ma1由2×9a1=3a1+6a1所以2x“(-94(-2,90-2,解得q=-1(q=1含去)又因为吗2+aq1-9=4,即aq(1+q)=4,所以aq=8,则a=a=(aq)…(q3)2=2
"20.【详解】(1)∵焦距为2√2,则c=√2,设A(x,y),B(x2y2),为弦AB的中点,根据中点坐标公式可得:x+x2=,y1+y2=又∵将其A(x,y1),B(x,y2)代入椭圆C:x+2=16x .+a162x2+a222=ab2将两式作差可得:b2(x+x2)(x1-x2)+a2(y+y2)(1-y2)=0,k==2=b(x+x)=32X-xa(i+ y∴a2=3b2-—①.∵a2-b2=c2--②由①②得:b2=1椭圆的标准方程为+y2=1(2)∵M,Q,N三点共线,OQ=OM+ON∴根据三点共线性质可得:-+-=1,则A=2设M(x1,y),N(x2y2),则x+x2=0,∴x=-2x2y=hr+m,将直线/和椭圆C联立方程消掉yx2+3y2=3可得:(1+3k2)x2+6kmx+3m2-3=0△>0→3k2-m2+1>0——①6km3m2-3根据韦达定理:x1+x2=1+321+3k代入x1=2,可得:x1+342~3m2-36km1+3k∴2、36k2m3m2-31+3k1+3k2,即(9(9m2-1)3k2=19m2-1≠0,m2≠∴3k21-m≥0—②9代入①式得m2+1>0,即+(1-m2)>0,9m2-19m2-1m(m2-1(9m2-1)<0,…
书面表达One possible version:NOTICEA sports meeting will be held in the playground of our school from next Thursday to FridayAs you know, the pressure of study is very heavy now, especially for those senior 3. So the purpose of the sports meeting is to letevery student get relaxed, as a result of which we students can live happily and heal thilyEveryone is welcome to take part in it. Those who perform excellently at the sports meeting will get prizes. But don' t take theresults so serously because taking part is more important than the result. Good luck to everyone!
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