Найти (x³³³³ + x³³³ +x³³ +x³ +1996)/100*(x² +x), если x² +x +1 =0 .
знаменатель дроби 100*(x² +x) =100*(x² +x+1 -1)= 100*(0 -1) = -100.
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x³³³³ + x³³³ +x³³ +x³ +1996= (x³³³³ -1) + (x³³³ -1) +(x³³-1) +(x³-1)+1992 =
((x³)¹¹¹¹ -1) + ((x³)¹¹¹ -1) +((x³)¹¹-1) +(x³-1)+1992 =
* * * 1111, 111,11,1 нечетные степени * * *
(x³ -1)*A(x) + (x³ -1)*B(x) +(x³-1)*C(x) +(x³-1)+1992 =
(x³ -1)*( A(x) +B(x) +*C(x) +1) +1992 =
(x -1)(x² +x +1)( A(x) +B(x) +*C(x) +1) +1992 =1992.
(x³³³³ + x³³³ +x³³ +x³ +1996)/100*(x² +x ) =1992/(-100) =-19,92
ответ : -19,92.
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Можно и так:
x² +x +1 =0 ;
D =( -1)² -4*1*1 = -3< 0⇒ корни кв уравнения комплексные числа
x₁ =(-1-i√3)/2 = -1/2 -(√3)/2 i ; * * * i = √(-1) * * *
x₂ =(-1+i√3)/2 = -1/2 +(√3)/2 i .
* * * z = r(cosφ+isinφ) тригонометрическая форма компл. числа) * * *
r₁ =r₂ =1 .
* * *модуль r₁ =√ ((-1/2)² +(-(√3)/2) )²) =1 ; r₂ =√ ((-1/2)² +(√3)/2)² ) * * *
φ₁ =arctq((-√3)/2)/(-1/2)) =π/3.
φ₁ =arctq((√3)/2)/(-1/2)) = -π/3.
x₁ =cos(π/3) + i*sin(π/3).
x₂ =cos(-π/3) + i*sin(-π/3) =cos(π/3) - i*sin(-π/3).
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x₁ =cos(π/3) + i*sin(π/3).
* * *( r(cosφ+isinφ) ) ^ n =(r^n)*(cos(nφ)+isin(nφ))_формула Муавра * * *
x₁³³³³ + x₁³³³ +x₁³³ +x₁³ +1996 = cos(1111π) + i*sin(1111π) +
cos(111π +i*sin(111π)+cos(11π)+ i*sin(11π) cos(π)+i*sin(π)+1996 =
-1-1-1-1+1996 =1992.
(x₁³³³³ + x₁³³³ +x₁³³ +x₁³ +1996)/(100(x²+x)) =1992/(-100) = -19,92.
(x₂³³³³ + x₂³³³ +x₂³³ +x₂³ +1996)/(-100) =1992/(-100) = -19,92.
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x₂ =cos(π/3) - i*sin(π/3).
(x₂³³³³ + x₂³³³ +x₂³³ +x₂³ +1996) = cos(1111π) - i*sin(1111π) +
cos(111π -i*sin(111π)+cos(11π)- i*sin(11π) cos(π)-i*sin(π)+1996 =
-1-1-1-1+1996 =1992.
(x₂³³³³ + x₂³³³ +x₂³³ +x₂³ +1996)/(100(x²+x)) =1992/(-100) = -19,92.