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2021年1月5日 星期二

106年新竹高商教甄-數學詳解

國立新竹高商106學年度第一次數學科教師甄選

一、填充題
x4+ax2+bx+c=(x1)3(xc)=(x33x2+3x1)(xc)x3=0=c3c=3{a=3c+3=6b=3c1=8(a,b,c)=(6,8,3)
{y=f(x)=x/3y=g(x)=cos(πx){g(x)=1x=2k,kZg(x)=1x=2k+1,kZ{f(x)1x3f(x)1x3x[3,3)x>03x<047

{¯AB=10sinθ¯BO=10cosθ¯BH=¯ABׯBO÷¯OA=100sinθcosθ÷10=10sinθcosθ¯AB+¯BO+¯BH=10(sinθ+cosθ)+10sinθcosθ=102sin(θ+45)+5sin(2θ)θ=45102+5
{n=1m=27n2m2=74=3此題並無特定解法!
{21x+22y+27z=50(1)22x+23y+28z=51(2)23x+24y+25z=52(3)(2)(1),(3)(2){x+y+z=1x+y3z=1{z=0x+y=1(1)21x+22(1x)=50x=28y=1x=29(x,y,z)=(28,29,0)
1x+1y=166x+yxy=166xy66(x+y)=0(x66)(y66)=612612=212×312(1+12)(1+12)=169169
1,4,7,...,3k+1,...,2017673
limn1n2nk=116n2(4k)2=limn1nnk=11616(kn)2=101616x2dx=1041x2dx=π/204cos2θdθ(x=sinθdx=cosθdθ)=π/202cos(2θ)+2dθ=[sin(2θ)+2θ]|π/20=π
{a2+b2=4(c5)2+(d12)2=36{(a2+b2)(122+(5)2)(12a5b)2(a2+(b)2)((d12)2+(c5)2)(a(d12)b(c5))2{4×169(12a5b)24×36(adbc12a+5b)2{2612a5b2612+12a5bacbd12+12ab12+26acbd122638acbd38acbd38
{A+B+C+D=16A,B,C,DNA>B{A+B+C+D=12A,B,C,D{0,1,2,,12}A>BABC+D1011H211=C12112010H210=C11101200H20=C10219H29=C1091110H20=C10651H21=C21750H20=C10=6n=12nk=1k=6n=1n(2n+1)=6n=12n2+6n=1n=282+21=303
1,sinθ,3cosθx3+ax+b=0{1+sinθ+3cosθ=0(1)sinθ+3sinθcosθ+3cosθ=a(2)3sinθcosθ=b(3)(1)sinθ+3cosθ=112sinθ+32cosθ=12sin(θ+30)=12θ=150(0<θ<π)(2)(3){a=123232332=74b=3232=34(a,b)=(74,34)
{A(0,2)B(2,0)C(0,2)D(2,0)1ABCDm2+n2=22+(2)2=8
C,F,D{¯AC=¯AD=a+2¯BC=¯BF=a2¯ED=¯EF=1{¯AB=4¯AE=a+1¯BE=a1ABE¯AB2=¯AE2+¯BE242=(a+1)2+(a1)2a2=7a=7

a+b+c33abc133abc3abc13a2+b2+b2+c2+c2+a233a2+b2b2+c2c2+a2332ab2bc2ca=323abca2+b2+b2+c2+c2+a2323abc2
{1000a+100b+10c+d1000d+100c+10b+a1000d+100c+10b+a(1000a+100b+10c+d)=4725999d+90c90b999a=4725111(da)10(bc)=525{da=5bc=3{(d,a)=(6,1),(7,2),(8,4),(9,5)(b,c)=(3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)4×7=28
f(93)=902f(90)=902(872f(87))=902872+f(87)=902872+842f(81)=902872+842+302f(30)=93f(30)=902872+842+30293=10k=0(30+6k)210k=0(33+6k)2+93293=10k=0(36k189)+93293=36×55189×11+93293=4497
:f(x)=(x+1)(x2+(a1)x+(ba+1))+cb+a1limx1f(x)x+1=3{f(x)=(x+1)g(x)g(1)=3g(x)=x2+(a1)x+(ba+1)g(1)=1+1a+ba+1=3b=2a(1)y=f(x)f(x)=3x2+2ax+b=004a212b0(1)4a212(2a)04a(a6)00a6
log2yx2=My=2Mx2{y=2Mx22x+y+2=0M2x+2Mx2+2=0:2282M=02M=12M=1
Γ:y2=4cx{A(α24c,α)B(β24c,β)C(γ24c,γ){G(α2+β2+γ212c,α+β+γ3)ABm1=βα(β2α2)/4c=4cβ+αBCm2=γβ(γ2β2)/4c=4cγ+βCAm3=αγ(α2γ2)/4c=4cα+γtan60=3=m2m11+m1m2=m3m21+m2m3=m1m31+m1m34cγ+β4cβ+α1+16c2(α+β)(β+γ)=4cα+γ4cβ+γ1+16c2(α+γ)(β+γ)=4cα+β4cα+γ1+16c2(α+β)(α+γ)=34c(αγ)16c2+(α+β)(β+γ)=4c(βα)16c2+(β+γ)(γ+α)=4c(γβ)16c2+(α+β)(α+γ)=34c(αγ)+4c(βα)+4c(γβ)48c2+(α+β)(β+γ)+(β+γ)(γ+α)+(γ+α)(α+β)=30048c2+(α+β)(β+γ)+(β+γ)(γ+α)+(γ+α)(α+β)=048c2+α2+β2+γ2+3(αβ+βγ+γα)=0αβ+βγ+γα=16c2(α2+β2+γ2)/3G(α2+β2+γ212c=x,α+β+γ3=y)(α+β+γ)2=9y2=α2+β2+γ2+2(αβ+βγ+γα)=12cx+2(αβ+βγ+γα)=12cx+2(16c2(α2+β2+γ2)/3)=12cx+2(16c24cx)=4cx32c29y2=4cx32c2


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