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anem2017
@anem2017
July 2022
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Помогите решить, 1,3,4
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MizoriesKun
8,4*1,3 12*0,7*1,3
------------- = -------------------- = 12*1.3 =15,6
0,7 0,7
2√5=√(2²*5)=√20
5√2=√(5²*2)=√50
6=√36
⇒ √20,√36,√50 ⇒
2√5, 6, 5√2
х²-5х=14
х²-5х-14=0
D=25+56=81 √D=9
x₁=(5+9)/2=7
x₂=(5-9)/2=-2
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Answers & Comments
------------- = -------------------- = 12*1.3 =15,6
0,7 0,7
2√5=√(2²*5)=√20
5√2=√(5²*2)=√50
6=√36
⇒ √20,√36,√50 ⇒2√5, 6, 5√2
х²-5х=14
х²-5х-14=0
D=25+56=81 √D=9
x₁=(5+9)/2=7
x₂=(5-9)/2=-2