Если
1/(х+4)+1/(х-5)=1/(х-2)+1/(х-4) , то
[(х-5)+(х+4)]/[(х-5)·(х+4)]=[(х-4)+(х-2)]/[(х-2)·(х-4)]
(2x-1)/(x²-x-20)=(2x-6)/(x²-6x+8)
(2x-1)·(x²-6x+8)=(2x-6)(x²-x-20)
2x³-13x²+22x-8=2x³-8x²-34x+120
-13x²+22x-8=-8x²-34x+120
5x²-56x+128=0
D=56²-4·5·128=3136-2560=576=24²
x1=(56-24)/10=32/10=16/5 x2=(56+24)/10=80/10=8
проверка - надо подставить в 1/(х+4)+1/(х-5)=1/(х-2)+1/(х-4)
x1=16/5
1/((16/5)+4)+1/((16/5)-5) = 1/((16/5)-2)+1/((16/5)-4)
10/72 -10/18=-30/72=-10/24=-5/12 5/6 -5/4=10/12-15/12=-5/12 верно
x2=8
1/(8+4)+1/(8-5) = 1/(8-2)+1/(8-4)
1/12+1/3=5/12 1/6+1/4 =(2+3)/12 = 5/12 верно
ответ: x1=16/5 x2=8