1
Sin(π/4+π/4)+2sin(π/4-π/4)+4cosπ/4+2cos(π/4+3π/4)=
=sinπ/2+2sin0+4cosπ/4+2cosπ=1+2*0+4*√2/2+2*(-1)=1+0+2√2-2=2√2-1
2
sin(π/4+3π/4)+2sin(π/4-3π/4)+4cos3π/4+2cos(3π/4+3π/4)=
=sinπ+2sin(-π/2)-4cosπ/4+2cos3π/2=0+2*(-1)-4*√2/2+2*0=-2-2√2
4) =8x³y-4xy-x³y=7x³y-4xy.---- степень 4
5)=4x³y-3xy-4x³y+6= -3xy+6--- степень 2
<em>Ответ:</em>
Т.к., Две стороны, лежащие напротив меньших углов - это катеты, то за т.Пифагора
третья сторона=√(√131)²+(√158)²=√289=17
|5x + 3| = |3 - x|
5x + 3 = 3 - x или 5x + 3 = - (3 - x)
5x + x = 3 - 3 5x + 3 = - 3 + x
6x = 0 5x - x = - 3 - 3
x₁ = 0 4x = - 6
x₂ = - 1,5
Lim┬(x→0)〖(1+x-x^2)/(〖2x〗^2+5x+4)〗 =
=lim┬(x→0)〖(1+x-x^2):x^2/(〖2x〗^2+5x+4):x^2〗=
=lim┬(x→0)〖(1/x^2+x/x^2-x^2/x^2)/(4x^2/x^2+5x/x^2+4/x^2)〗=
= lim┬(x→0))〖(1/x^2+1/x-1)/(4+5/x+4/x^2)〗=-1/4