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abikozei123
@abikozei123
July 2022
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Помогите решить два номера по алгебре
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Hikaru007
Решение на фотографии
2 votes
Thanks 1
evachka15
168.
а) m*√x/√x*√x = m√x/x
б) 1*√2/√2*√2 = √2/2
в) 3*√c/5√c*√c = 3√c/5c
г) a*√3/2√3*√3 = a√3/2*3 = a√3/6
д) 5*√15/4√15*√15 = 5√15/4*15 = 5√15/60 = √15/12
169.
а) 4*(√3-1)/(√3+1)*(√3-1) = 4(√3-1)/√3-1 = 4
в) 1*(√х+√у)/(√х+√у)*(√х-√у) = √х+√у/х-у
д) 33*(7+3√3)/(7-3√3)*(7+3√3) = 33*(7+3√3)/49-27 = 3*(7+3√3)/2 = 21+9√3/2
б) 1*(1+√2)/(1-√2)*(1+√2) = 1+√2/1-2 = 1+√2/-1 = -1-√2
г) а(√а-√b)/(√a-√b)*(√a+√b) = a(√a-√b)/a-b
e) 15*(2√5-5)/(2√5+5)*(2√5-5) = 15*(2√5-5)/20-25 = -6√5 + 15
2 votes
Thanks 2
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Answers & Comments
Решение на фотографии
а) m*√x/√x*√x = m√x/x
б) 1*√2/√2*√2 = √2/2
в) 3*√c/5√c*√c = 3√c/5c
г) a*√3/2√3*√3 = a√3/2*3 = a√3/6
д) 5*√15/4√15*√15 = 5√15/4*15 = 5√15/60 = √15/12
169.
а) 4*(√3-1)/(√3+1)*(√3-1) = 4(√3-1)/√3-1 = 4
в) 1*(√х+√у)/(√х+√у)*(√х-√у) = √х+√у/х-у
д) 33*(7+3√3)/(7-3√3)*(7+3√3) = 33*(7+3√3)/49-27 = 3*(7+3√3)/2 = 21+9√3/2
б) 1*(1+√2)/(1-√2)*(1+√2) = 1+√2/1-2 = 1+√2/-1 = -1-√2
г) а(√а-√b)/(√a-√b)*(√a+√b) = a(√a-√b)/a-b
e) 15*(2√5-5)/(2√5+5)*(2√5-5) = 15*(2√5-5)/20-25 = -6√5 + 15