пусть n- искомое число,тогда следующее за ним число: n+1. Составляем уравнение:
(n+1)^2-n^2=65
n^2+2n+1-n^2=65
2n+1=65
2n=64
n=32
2сos(5x/2)cos(x/2)-2sin(5x/2)cos(5x/2)=0
2cos(5x/2)*(cos(x/2)-sin(5x/2))=0
cos(5x/2)=0⇒5x/2=π/2+πn⇒x=π/5+2πn/5,n∈z
cos(x/2)-sin(5x/2)=0
cos(x/2)-cos(π/2-5x/2)=0
-2sin(3x/2-π/4)sin(-x+π/4)=0
2sin(3x/2-π/4)sin(x-π/4)=0
sin(3x/2-π/4)=0⇒3x/2-π/4=πk⇒3x/2=π/4+πk⇒x=π/6+2πk/3,k∈z
sin(x-π/4)=0⇒x-π/4=πm⇒x=π/4+πm,m∈z
1) 6 x - 10,2 = 4 x = - 2,2 x/0,2
2 x = 8 X = 4
2) 15 - ( 3 x - 3) = 5 - 4 x - 3 x + 4 x = 5 - 15 - 3 x = - 13
3) 2 x - 1 + 1 = 9
2 x = 9 + 1 - 1
2 x = 9
X = 4, 5