7√28 - √80 - 2√63 + 3√45 = 7√7 * √ 4 - √16* √5 -2√7 *√9 + 3√5 * √9 =
14√7-4√5-6√7+9√5 = 8√7 + 5√5 (вроде так)
Ctg3x = tg5x
cos3x/sin3x = sin5x/cos5x
ОДЗ:
sin3x ≠ 0
3x ≠ πn, n ∈ Z
x ≠ πn/3, n ∈ Z
cos5x ≠ π/2 + πn, n ∈ Z
x ≠ π/10 + πn/5, n ∈ Z
sin5x/cos5x - cos3x/sin3x = 0
(sin5sin3x - cos5xcos3x)/sin3xcos5x = 0
cos5xcos3x - sin5xsin3x = 0
cos(5x + 3x) = 0
cos8x = 0
8x = π/2 + πn, n ∈ Z
x = π/16 + πn/8, n ∈ Z
Если построить графики функций y = sin3x, y = cos5x, y = cos8x, то можно увидеть, что в общих точек у графиков при пересечении оси Ox нет.
Ответ:x = π/16 + πn/8, n ∈ Z.
1)α-β=102⁰;⇒α=102⁰+β;
α+β=180⁰⇒102⁰+β+β=180⁰⇒102⁰+2β=180⁰
2β=180⁰-102⁰=78⁰;
β=78⁰/2=39⁰; α=102⁰+39⁰=141⁰;
2)180⁰-124⁰=56⁰
A
[(x+3)(x-1)+(x+1)√(x+3)(x-3)]/[(x-3)(x+1)+(x-1)√(x-3)(x+3)]=
=√(x+3)[(x-1)√(x+3)+(x+1)√(x-3)]/√(x-3)[(x+1)√(x-3)+(x-1)√(x+3)]=
=√(x+3)/√(x-3)=√[(x+3)/(x-3)]
б
[(t-3)(t+2)+(t+3)√(t-2)(t+2)]/[(t+3)(t-2)-(t-3)√(y-2)(t+2)]=
=√(t+2)[(t-3)√(t+2)-(t+3)√(t-2)]/√(t+2)[(t+3)√(t-2)-(t-3)√(t+2)]=
=-√(t+2)/√(t-2)=-√[(t+2)/(t-2)]