Ответ(ы) на вопрос:
Гость886.
(3/(25 -x²) +1/(x²-10x+25) )*(x-5)²/2 +3x/(x+5) =
(1/(x-5)² -3/(x-5)(x+5))*(x-5)²/2 +3x/(x+5) =1/2 -3(x-5)/2(x+5) +3x/(x+5) =
1/2 +(2*3x -3(x-5))/2(x+5) =1/2 +3(x+5)/2(x+5)= 1/2+3/2 =2.
иначе
(3/(25 -x²) +1/(x²-10x+25) )*(x-5)²/2 +3x/(x+5) =
(3/(5²-x²) +1/(x²-2x*5+5²) )*(x-5)²/2 +3x/(x+5) =
(3/(5-x)(5+x) +1/(x-5)² )*(x-5)²/2 +3x/(x+5) =
(-3/(x -5)(x+5) +1/(x -5)² )* (x-5)²/2 +3x/(x+5)=
(-3x +15 +x+5)/(x -5)²(x+5) * (x-5)²/2+3x/(x+5) =
2(10-x)/(x-5)²(x+5)*(x-5)²/2 +3x/(x+5)=
(10-x)/(x+5)+ 3x/(x+5)=(10-x+3x) /(5+x) =2(x+5)/(x+5) =2.
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887.
{a+aq +aq²=7 ;a² +a²q² +a²q⁴=21.⇔{a(1+q +q²)=7 ;a²(1 +q² +(q²)²)=21.⇔
{a(1+q +q²)=7 ;a²(1 +2q² +(q²)² -q²)=21.(прибили и вычитали q²)⇔
{a(1+q +q²)=7 ;a²((1 +q²)² -q²)=21.⇔
{a(1+q +q²)=7 ;a(1 +q² -q)*a(1 +q² +q)=21.⇔
{a(1+q +q²)=7 ; a(1 +q² -q) =3.⇒
(1+q +q²)/(1 +q² -q) =7/3.⇔3(q²+q+1) =7(q²-q+1) ;
4q²-10q +4 =0 ;
2q²-5q +2 =0 ;
q² -(2+1/2)q +1 =0 ⇒[q =2 ; q =1/2.
q =2 ⇒ a =7/(q² +q+1) =1.
Члены прогрессии: 1 ; 2 ; 4.
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q =1/2 ⇒ a =7/(q² +q+1) =4.
Члены прогрессии: 4 ;2; 1.
(в обратном порядке первой последовательности).
ответ : {1;2;4} или {4;2;1}.
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