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nikitaqwerty1
@nikitaqwerty1
October 2021
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Помогите пожалуйста решите плиз
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prostomama1
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2(х-у)/у ^ 3у*2/(х-у)(х+у)=6у*2/у(х+у)=3/(х+у)
(15а*2-15а*2+10а)/3а-2=10а/(3а-2)
(6с*2-9-6с*2)/3+2с=-9/(3+2с)
(2у*2-2у*2+16у)/у-8=16у/(у-8)
9а-3а*2-9а/(а+3)=-3а*2/(а+3)
5а*2+5а-3-5а*2/(а+1)=(5а-3)/(а+1)
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nikitaqwerty1
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nikitaqwerty1
помогите пожалуйста
ТатМих
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Answers & Comments
2(х-у)/у ^ 3у*2/(х-у)(х+у)=6у*2/у(х+у)=3/(х+у)
(15а*2-15а*2+10а)/3а-2=10а/(3а-2)
(6с*2-9-6с*2)/3+2с=-9/(3+2с)
(2у*2-2у*2+16у)/у-8=16у/(у-8)
9а-3а*2-9а/(а+3)=-3а*2/(а+3)
5а*2+5а-3-5а*2/(а+1)=(5а-3)/(а+1)
Verified answer
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