Помогите с 886, пожалуйста!

886. (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. ------- 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. --- 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}.