Решите пожалуйста 36.8 (б) и 36.10 (а и г) очень надо! !!

36.8.b  2x-1≠0        ⇒  x≠ 0,5 ; x²-2x-6 ≠0   ⇒   x≠ 1-√7  ;  x≠ 1+√7  (x+2-4)/(2x-1)= (-x+2)/(x²-2x-6) (x-2)/(2x-1) +((x-2)/(x²-2x-6)=0 (x-2)·[1/(2x-1) + 1/(x²-2x-6)] = 0    x-2 =0   ⇒  x=2  1/(2x-1) +1/(x²-2x-6) =0 (x²-2x-6)+(2x-1)=0  x²-7=0     ⇒x=+/-√7 Ответ:  2; -√7; √7 36.10.a (x+1)/[x²(x-3) +(x-3)] +1/[(x²+1)·(x²-1)] = (x-2)/[(x²(x-3) - (x-3)] (x+1)/[(x-3)(x²+1)] + 1/[(x²+1)(x²-1)] = (x-2)/[(x-3)(x²-1)]          общ.  зн.   (x-3)(x²+1)(x²-1)  ⇒ (x+1)·(x²-1) + (x-3) = (x-2)·(x²+1) x³+x²-x-1 +x-3 = x³-2x²+x-2 3x²-x-2=0 x= [1+/-√(1+4·3·2)]/2·3 = [1+/-5]/6  x1=-4 :  x2= 26/6= 13/3 Ответ:  -4; 13/3 36.10.г. 5/[(x³+1) -2x(x+1)] - 2/[(x³-1)- 4x(x-1)] =1/(x²-1) 5/[(x+1)(x²-x+1 -2x)] - 2/[(x-1)(x²+x+1-4x)] = 1/[(x+1)(x-1)]  5/[(x+1)(x²-3x+1)] - 2/[(x-1)(x²-3x+1)] = 1/[(x+1)(x-1)]  5(x-1) - 2(x+1) = x²-3x+1 x²-6x+8 = 0 x1=2 :  x2 = 4 Ответ: 2; 4