e ơi e nên tải tài liệu của võ quốc bá cẩn đi 

Mai mình nộp rồi! giúp mình với!mình tìm ở tất cả các trang mà không thấy! i need help!

dễ quá

\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{d}{b}=\frac{c}{a}\Leftrightarrow\frac{d^{2016}}{b^{2016}}=\frac{c^{2016}}{a^{2016}}=\frac{c^{2016}-d^{2016}}{a^{2016}-b^{2016}}=\frac{c^{2016}+d^{2016}}{a^{2016}+b^{2016}}\)

(áp dụng tính chất dãy tỉ số bằng nhau)

Suy ra \(\frac{a^{2016}+b^{2016}}{a^{2016}-b^{2016}}.\frac{c^{2016}-d^{2016}}{c^{2016}+d^{2016}}=\frac{a^{2016}+b^{2016}}{c^{2016}+d^{2016}}.\frac{c^{2016}-d^{2016}}{a^{2016}-b^{2016}}\)

\(=\frac{b^{2016}}{d^{2016}}.\frac{d^{2016}}{b^{2016}}=1\)

\(P=\frac{2a+3b+3c-1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c+1}{2017+c}\)

\(=\frac{6047-a}{2015+a}+\frac{6048-b}{2016+b}+\frac{6049-c}{2017+c}\)

\(=\frac{8062}{2015+a}+\frac{8064}{2016+b}+\frac{8066}{2017+c}-3\)

\(\ge\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{2015+2016+2017+a+b+c}-3=\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{8064}-3\)

Dấu = xảy ra khi ....

\(A=\frac{2016a}{ab+2016a+2016}+\frac{b}{bc+b+2016}+\frac{c}{ac+c+1}\)

\(A=\frac{2016a}{ab+2016a+abc}+\frac{b}{bc+b+2016}+\frac{bc}{abc+bc+b}\)

\(A=\frac{2016a}{a\left(b+2016+bc\right)}+\frac{b}{bc+b+2016}+\frac{bc}{2016+bc+b}\)

\(A=\frac{2016}{b+2016+bc}+\frac{b}{bc+b+2016}+\frac{bc}{2016+bc+b}\)

\(A=\frac{2016+b+bc}{2016+b+bc}=1\)

Thay : 2016 = abc

ta có :

\(A=\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(A=\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)

\(A=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)

\(A=\frac{ac+c+1}{ac+c+1}\)

\(A=1\)

vậy \(A=\frac{2016.a}{ab+2016.a+2016}+\frac{b}{bc+b+2016}+\frac{c}{ac+c+1}=1\)

Chúc bạn học tốt !