60 reputation
5
bio website diendan.hocmai.vn
location
age 16
visits member for 1 year, 6 months
seen Jan 28 at 2:27

Dec
27
revised Integrals: $I=\int_{1}^{e}\frac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$
added 2 characters in body
Dec
27
accepted Integrals: $I=\int_{1}^{e}\frac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$
Dec
27
answered Integrals: $I=\int_{1}^{e}\frac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$
Dec
27
awarded  Editor
Dec
27
revised Integrals: $I=\int_{1}^{e}\frac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$
edited body
Dec
27
asked Integrals: $I=\int_{1}^{e}\frac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$
Dec
26
comment Does anyone know of any additive periodic functions?
We see at [enter link description here][1] [1]: math.stackexchange.com/questions/43964/…
Dec
26
awarded  Supporter
Dec
25
awarded  Teacher
Dec
25
accepted Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$
Dec
25
answered Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$
Dec
23
asked Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$
Dec
22
comment $(\cot \alpha)^{\cos 2\alpha} \ge \frac{1}{\sin 2\alpha}$
I don't know. Can you help me? Thank you.
Dec
22
comment $(\cot \alpha)^{\cos 2\alpha} \ge \frac{1}{\sin 2\alpha}$
I am VietNamese ; so I bad good at English.
Dec
21
asked $(\cot \alpha)^{\cos 2\alpha} \ge \frac{1}{\sin 2\alpha}$
Dec
21
accepted $\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$
Dec
21
comment $\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$
Thank you very much
Dec
21
awarded  Scholar
Dec
20
awarded  Student
Dec
20
asked $\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$