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′=4x3−2(m+1)x=0⇔2x[2x2−(m+1)]=0⇔[x=0⇒y=mx2=m+12(1)Dehamsoco3CT⇒(1)co2nghiempb≠0⇒m+12>0⇔m>−1(1)⇔x=±√m+12⇒y=(m+1)24−(m+1)m+12+my=−(m+1)24+m⇒A(0;m);B(√m+12;−(m+1)24+m);C(−√m+12;−(m+1)24+m)⇒ΔABCcantaiA.a)ΔABCvuongcantaiA⇒→AB.→AC=0→AB=(√m+12;−(m+1)24);→AC=(−√m+12;−(m+1)24)→AB.→AC=0⇔−m+12+(m+1)416=0⇔(m+1)4−8(m+1)=0⇔(m+1)[(m+1)3−8]=0⇔[m+1=0m+1=2⇔[m=−1(loai)m=1(tm)Vaym=1.b)ΔABCdeu⇔AB2=BC2⇔m+12+(m+1)416=4.m+12⇔(m+1)416=3m+12⇔(m+1)4=24(m+1)⇔(m+1)[(m+1)3−24]=0⇔[m=−1(loai)m+1=3√24⇔m=3√24−1(tm)c)ΔABCco1goc1200⇒ˆA=1200⇒cosA=→AB.→ACAB.AC=(m+1)[(m+1)3−8]m+12+(m+1)416=−12⇔−(m+1)[(m+1)3−8]=m+12+(m+1)416⇔−[(m+1)3−8]=12+(m+1)316(Dom≠−1)⇔−16[(m+1)3−8]=8+(m+1)3⇔17(m+1)3=120⇔(m+1)3=12017⇔m+1=3√12017⇔m=3√12017−1(tm) d)SABC=12d(A;BC).BCPTBC:y=−(m+1)24+m⇒d(A;BC)=|m+(m+1)24−m|=(m+1)24BC=2√m+12⇒SABC=12.(m+1)24.2√m+12=8⇔√m+125=64⇔√m+12=5√64⇔m+12=(5√64)2⇔m+1=2(5√64)2⇔m=2(5√64)2−1(tm)e)ApdungCT:R=abc4Sf)ApdungCT:r=Sp