26
(1-√2)/((1+√2)(1-√2))+(√2-√3)/((√2+√3)(√2-√3))+9√3-2)/((√3+2)(√3-2))=
=(1-√2)/(1-2)+(√2-√3)/(2-3)+(√3-2)/(3-4)=
=-1+√2-√2+√3-√3+2=1
27
2sin(x/2)cos(x/2)/(2cos²(x/2))=sin(x/2)
si(x/2)/cos(x/2)-sin(x/2)=0
cos(x/2)≠0⇒x/2≠π/2+πn,n∈z⇒x≠π+2πn,n∈z
sin(x/2)*(1-cos(x/2))=0
sin(x/2)=0⇒x/2=πk,k∈z⇒x=2πk,k∈z
π≤2πk≤2π
1≤2k≤2
1/2≤k≤1
k=1⇒x=2π
cos(x/2)=1⇒x/2=2πm,m∈z⇒x=4πm,m∈z
π≤4πm≤2π
1≤4m≤2
1/4≤m≤1/2
нет решения на промежутке
Ответ x=2π