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Question Number 180542 by a.lgnaoui last updated on 13/Nov/22
Resoudre dans R  1)  a+b+c=2  a^2 +b^2 +c^2 =6  (1/a)+(1/b)+(1/c)=(1/2)    2)  x^2 +xy+y^2 =3  y^2 +yz+z^2 =7  z^2 +zx+x^2 =13
ResoudredansR1)a+b+c=2a2+b2+c2=61a+1b+1c=122)x2+xy+y2=3y2+yz+z2=7z2+zx+x2=13
Answered by mr W last updated on 13/Nov/22
1)  a+b+c=2  (a+b+c)^2 =2^2 =4  a^2 +b^2 +c^2 +2(ab+bc+ca)=4  ⇒ab+bc+ca=((4−6)/2)=−1  ((ab+bc+ca)/(abc))=(1/2)  ⇒abc=2(−1)=−2  a,b,c are roots of  x^3 −2x^2 −x+2=0  (x−2)(x−1)(x+1)=0  ⇒a,b,c=(−1,1,2)
1)a+b+c=2(a+b+c)2=22=4a2+b2+c2+2(ab+bc+ca)=4ab+bc+ca=462=1ab+bc+caabc=12abc=2(1)=2a,b,carerootsofx32x2x+2=0(x2)(x1)(x+1)=0a,b,c=(1,1,2)
Commented by a.lgnaoui last updated on 13/Nov/22
thanks
thanks
Answered by mr W last updated on 14/Nov/22
2)  a=(√7),b=(√(13)),c=(√3)  Δ=((√(((√(13))+(√7)+(√3))(−(√(13))+(√7)+(√3))((√(13))−(√7)+(√3))((√(13))+(√7)−(√3))))/4)=((5(√3))/4)  x=±((4(√6)×((5(√3))/4)+3(√2)(−7+13+3))/(6(√(13+7+3+4(√3)×((5(√3))/4)))))  x=±((15(√2)+3(√2)(−7+13+3))/(6(√(13+7+3+15))))=±(7/( (√(19))))  y=±((15(√2)+3(√2)(7−13+3))/(6(√(13+7+3+15))))=±(1/( (√(19))))  z=±((15(√2)+3(√2)(7+13−3))/(6(√(13+7+3+15))))=±((11)/( (√(19))))
2)a=7,b=13,c=3Δ=(13+7+3)(13+7+3)(137+3)(13+73)4=534x=±46×534+32(7+13+3)613+7+3+43×534x=±152+32(7+13+3)613+7+3+15=±719y=±152+32(713+3)613+7+3+15=±119z=±152+32(7+133)613+7+3+15=±1119

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