Tìm m để đths y=x^4 - (m+1)x^2 +m sao cho : a, Tam giác ABC vuông cân b, Tam giác ABC đều c, Tam giác ABC coa 1 góc = 120 độ d, Diện tích tam giác ABC=8 e, Bán kính đt ngoại tiếp=1 f, Bán kính đt nội tiếp =1

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y=x4(m+1)x2+my=4x32(m+1)x=02x[2x2(m+1)]=0[x=0y=mx2=m+12(1)Dehamsoco3CT(1)co2nghiempb0m+12>0m>1(1)x=±m+12y=(m+1)24(m+1)m+12+my=(m+1)24+mA(0;m);B(m+12;(m+1)24+m);C(m+12;(m+1)24+m)ΔABCcantaiA.a)ΔABCvuongcantaiAAB.AC=0AB=(m+12;(m+1)24);AC=(m+12;(m+1)24)AB.AC=0m+12+(m+1)416=0(m+1)48(m+1)=0(m+1)[(m+1)38]=0[m+1=0m+1=2[m=1(loai)m=1(tm)Vaym=1.b)ΔABCdeuAB2=BC2m+12+(m+1)416=4.m+12(m+1)416=3m+12(m+1)4=24(m+1)(m+1)[(m+1)324]=0[m=1(loai)m+1=324m=3241(tm)c)ΔABCco1goc1200ˆA=1200cosA=AB.ACAB.AC=(m+1)[(m+1)38]m+12+(m+1)416=12(m+1)[(m+1)38]=m+12+(m+1)416[(m+1)38]=12+(m+1)316(Dom1)16[(m+1)38]=8+(m+1)317(m+1)3=120(m+1)3=12017m+1=312017m=3120171(tm) d)SABC=12d(A;BC).BCPTBC:y=(m+1)24+md(A;BC)=|m+(m+1)24m|=(m+1)24BC=2m+12SABC=12.(m+1)24.2m+12=8m+125=64m+12=564m+12=(564)2m+1=2(564)2m=2(564)21(tm)e)ApdungCT:R=abc4Sf)ApdungCT:r=Sp

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