(х²-5)/(х-1)(х-2) - (х+3)/(х-1) = (2х+2)/(х-2)
ОДЗ: х≠1, х≠2
х²-5-(х+3)(х-2)=(2х+2)(х-1)
х²-5-х²+2х-3х+6=2х²-2х+2х-2
2х²+х-3=0
D=1+24=25
х₁ = -1-5 / 4 = -1,5
х₂ = -1+5 / 4 = 1 не принадлежит ОДЗ
Ответ: х = -1,5
(cos²a-tg²a*cos²a)/(2cos2a*sina)=(cos²a-sin²a)/(2cos2a*sina)=
=cos2a/(2cos2a*sina)=1/(2sina)
Решение .
1) Sin (-150) = sin (-90-60) = - cos(60)=-0,5
Ответ : Sina =-0,5.
(5x/(x-10) +20x/(x²-20x +100) :(4x-24)/(x² -100) -25x/(x-10) =
((5x(x-10) +20x) / (x-10)²)*(x-10)(x+10)/4(x-6) - 25x/(x-10) =
(5x(x-6) / (x-10)²)*(x-10)(x+10)/4(x-6) - 25x/(x-10) =
(5x / (x-10))*(x+10)/4 - 25x/(x-10) =5x(x+10) / 4(x-10) - 25x/(x-10) =
=(5x(x+10) -4*25x) / 4(x-10)=5x(x+10 -20)/4(x-10)=5x(x-10)/4(x-10) = 5x/4. || для x≠6 ; x≠10 ||
---------
(4x-24)/(a² -100) ? заменено на (4x-24)/(x² -100)