Cho $a^2+b^2=c^2+d^2=1$ ;$a+c=\frac{\sqrt{2}}{2}$; $b+d=\frac{\sqrt{6}}{2}$ .Tính $ad+bc$.
Edited by hoangquochung3042002, 24-05-2017 - 12:49.
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Posted 24-05-2017 - 20:46
khgisongsong
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- adteams, Tea Coffee and duylax2412 like this
$\frac{(x!)^2.(-1)^x+1}{2x+1}\in Z $ (với $x\in N)<=>2x+1$ là số nguyên tố
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