(√6-√3)(2√6-√3)=12-√18-2√18+3=12-3√3-2·3√2+3=12-3√2-6√2+3=15-9√2.
6(1-cos²x)+5cosx=0
6cos²x-5cosx-6=0
cosx=a
6a²-5a-6=0
D=25+144=169
a1=(5-13)/12=-2/3
cosx=-2/3
x=π-arccos2/3+2πn,n∈z
x=-4π+arccos2/3∈[-5π;-7π/2]
a2=(5+13)/12=3/2⇒cosx=1,5>1 нет решения
4*(1+cos2x)²/4 -1 =cos2x
(1+cos2x)² -(1+cos2x)=0
(1+cos2x)(1+cos2x-1)=0
(1+cos2x)=0⇒cos2x=-1⇒2x=π+2πn⇒x=π/2+πn⇒x=-π/2∈[-5π/6;π/6]
(1+cos2x-1)=0⇒cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2⇒x=-3π/4;-π/4∈[-5π/6;π/6]
3/(X+2)-2/(X+3)=1/(X+1)
[3(X+3)-2(X+2)]/[(X+2)(X+3)]=1/(X+1)
[X+5]/[(X+2)(X+3)]=1/(X+1) ⇒ (X+5)(X+1)=(X+2)(X+3) ⇒
X²+6X+5=X²+5X+6 ⇒X=1
ПРОВЕРКА
3/(1+2)-2/(1+3)=1/(1+1) ВЕРНО
Ответ: сумма равна 5*(-3)-10+5*2-10=-25+10-10=-25.
Объяснение: