Matematică, întrebare adresată de anonim98, 9 ani în urmă

Fie ecuatia x²-4x+7=0 si x₁, x₂ solutiile ei. Sa se calculeze:

(x₁⁵+4x₁³+1)/x₁⁴ + (x₂⁵+4x₂³+1)/x₂⁴.

Va rog ajutati-ma!
Multumesc anticipat! :)

Răspunsuri la întrebare

Răspuns de anamariaangel
17
 x^{2} -4x+7=0
Δ= b^{2}-4ac=(-4) ^{2}-4*7=16-28=-12
 x_{1} = \frac{-b-i \sqrt{-delta} }{2a} = \frac{4-i \sqrt{12} }{2} = \frac{4-2i \sqrt{3} }{2} = \frac{2(2-i \sqrt{3}) }{2} =2-i\sqrt{3}
 x_{2} = \frac{-b+i \sqrt{-delta} }{2a} = \frac{4+i \sqrt{12} }{2} = \frac{4+2i \sqrt{3} }{2} = \frac{2(2+i \sqrt{3}) }{2} =2+i\sqrt{3}

 x_{1}  ^{5}= (2-i \sqrt{3} ) ^{5}=(2-i \sqrt{3} ) ^{3}*(2-i \sqrt{3} ) ^{2}=[2 ^{3} +3*4*(-i \sqrt{3} )+3*2*(-i \sqrt{3}) ^{2} +(-i \sqrt{3}) ^{3}]*(4-4i \sqrt{3} +3i ^{2} )=(8-12i \sqrt{3}-18+3i \sqrt{3}  )*(4-4i \sqrt{3} )=(-10-9i  \sqrt{3} )*(1-4 \sqrt{3})= -10+40i \sqrt{3}-9i \sqrt{3}+36*3*i ^{2}   =-10+31i \sqrt{3} -108=-118+31i \sqrt{3}

 x_{1} ^{3}= (2-i \sqrt{3} ) ^{3}=2^{2}-3*4*(i \sqrt{3}  )+3*2*(-i \sqrt{3} ) ^{2} -(i \sqrt{3} ) ^{3} =8-12i \sqrt{3} -18+3 i\sqrt{3} =-10-9i \sqrt{3}

x_{1} ^{4}= (2-i \sqrt{3} ) ^{3}*(2-i \sqrt{3} )=(-10-9 i\sqrt{3} )*(2-i \sqrt{3} )=-20+10i \sqrt{3} -18i \sqrt{3} +27i ^{2} =-20-8i \sqrt{3}-27=-47-8i \sqrt{3}

x_{2} ^{3}= (2+i \sqrt{3} ) ^{3}=2 ^{3}+3*4* i \sqrt{3} +3*2*(i \sqrt{3} ) ^{2} +(i \sqrt{3} )^{3}=8+12 i \sqrt{3} +6*3i ^{2} +3i ^{3} \sqrt{3}=8+ 12 i \sqrt{3}-18-3i  \sqrt{3}=-10+9i  \sqrt{3}

x_{2} ^{4}= (2+i \sqrt{3} ) ^{3}*(2+i \sqrt{3} )=(-10+9i \sqrt{3})*(2+i \sqrt{3} ) =-20-10i \sqrt{3}+18i \sqrt{3}+27i ^{2} =-20+8i \sqrt{3}-27=-47+8i \sqrt{3}

x_{2} ^{5}= (2+i \sqrt{3} ) ^{4}*(2+i \sqrt{3} )=(-47+8i\sqrt{3})*(2+i \sqrt{3} )=-94-47i\sqrt{3}+16i\sqrt{3}+24i ^{2} =-94-31i\sqrt{3}-24=-118-31i\sqrt{3}

[tex] \frac{ x_{1} ^{5} +4x_{1} ^{3} +1}{x_{1} ^{4} } + \frac{ x_{2} ^{5} +4x_{2} ^{3} +1}{x_{2} ^{4} }= \frac{(2-i \sqrt{3} ) ^{5} +4(2-i \sqrt{3} ) ^{3}+1}{(2-i \sqrt{3} ) ^{4}} +\frac{(2+i \sqrt{3} ) ^{5} +4(2+i \sqrt{3} ) ^{3}+1}{(2+i \sqrt{3} ) ^{4}} [/tex]= \frac{-118+31i \sqrt{3}+4(-10-9i\sqrt{3})+1 }{-47-8i\sqrt{3}}+ \frac{-118-31i \sqrt{3}+4(-10+9i\sqrt{3})+1 }{-47+8i\sqrt{3}}=\frac{-118+31i \sqrt{3}-40-36i\sqrt{3}+1 }{-47-8i\sqrt{3}}+\frac{-118-31i \sqrt{3}-40+36i\sqrt{3}+1 }{-47-8i\sqrt{3}}\frac{-157-5i \sqrt{3} }{-47-8i\sqrt{3}}+\frac{-157+5i \sqrt{3}}{-47+8i\sqrt{3}}= \frac{(-157-5i \sqrt{3})*(-47+8i\sqrt{3})+(-157+5i \sqrt{3})*(-47-8i\sqrt{3})}{(-47-8i\sqrt{3})*(-47+8i\sqrt{3})} = \frac{(157*47-157*8i \sqrt{3}+47*5i \sqrt{3}-120)+(157*47+157*8i \sqrt{3}-47*5i \sqrt{3}+120)}{(-47) ^{2}-(8i \sqrt{3} ) ^{2} }  = \frac{7499-1021i \sqrt{3}+7499+1021i \sqrt{3} }{2209-192i ^{2} } = \frac{14998}{2401} =6,24



anonim98: Dar se poate daca se lucreaza in multimea numerelor complexe... Scuze, dar am uitat sa mentionez. :D
anamariaangel: :)
anamariaangel: * este inmultit; si i^2=-1
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