№1 (1+√15)(√3-√5) №2 (2√2/3 -5√3/8+4√3/2)*√2/3 - вопрос №7748199111535222353843223 от NaumYT 26.02.2023 22:49

Question

Алгебра: №1 (1+√15)(√3-√5) №2 (2√2/3 -5√3/8+4√3/2)*√2/3

NaumYT · Answer

№1 (1+√15)*(√3-√5)=√3-√5+√15*√3-√15*√5=√3-√5+√(5*3*3)-√(5*3*5)=
=√3-√5+3*√5-5*√3=√5*(-1+3)+√3*(1-5)=2*√5-4*√3
№2 (2*√(2/3)-5*√(3/8)+4*√(3/2))*√(2/3)
=(2*√2*√2)/(√3*√3)-(5*√3*√2)/(√8*√3)+(4*√3*√2)/(√2*√3)=
=(4/3)-(5*√6)/(√24)+4=(4/3)-(5*√6)/(√6*√4)+4=(4/3)-(5/2)+4=
=(4*2)/(3*2)-(5*3)/(2*3)+(4*6)/6=
=(8-15+24)/6=17/6