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