- Nhân cả tử và mẫu phân thức thứ nhất với a

- Nhân cả tử và mẫu phân thức thứ 2 với ac

- Thay abc =2016 ta có mẫu số chung là :

3ac - 4032 +2016a

- Rút gọn => đáp án : -1

\(P=\frac{2bc-2016}{3c-2bc+2016}-\frac{2b}{3-2b+ab}-\frac{4032-3ac}{3ac-4032+2016a}\)

Ta rút gọn từng biểu thức

\(+)\frac{2bc-2016}{3c-2bc+2016}=-1+\frac{3c}{3c-2bc+2016}\)

\(+)\frac{-2b}{3-2b+ab}=\frac{-2bc}{3c-2bc+abc}=\frac{-2bc}{3c-2bc+2016}\)

\(+)\frac{4032-3ac}{3ac-4032+2016a}=-1+\frac{2016a}{3ac-2abc+2016a}\)

\(=-1+\frac{2016}{3c-2bc+2016}\)

\(\Rightarrow P=-1\)

\(P=\dfrac{2bc-2016}{3c-2bc+2016}-\dfrac{2b}{3-2b+ab}+\dfrac{4032-3ac}{3ac-4032+2016c}\)
\(=\dfrac{2bc-abc}{3c-2bc+abc}-\dfrac{2b}{3-2b+ab}+\dfrac{2abc-3ac}{3ac-2abc+a^2bc}\)
\(=\dfrac{2b-ab}{3-2b+ab}-\dfrac{2b}{3-2b+ab}+\dfrac{2b-3}{3-2b+ab}\)
\(=\dfrac{2b-ab-2b+2b-3}{3-2b+ab}\)
\(=\dfrac{-3+2b-ab}{3-2b+ab}=-1\).

`(2bc-2016)/(3c-2bc+2016)`

`=(-(3c-2bc+2016)+3c)/(3c-2bc+2016)`

`=-1+(3c)/(3c-2bc+2016)`

`(2b)/(3-2b+ab)

`=(2bc)/(3c-2bc+abc)`

`=(2bc)/(3c-2bc+2016)`

`(4032-3ac)/(3ac-4032+2016a)`

`=(-(3ac-4032+2016a)+2016a)/(3ac-4032+2016a)`

`=-1+(2016a)/(3ac-2abc+2016a)`

`=-1+(2016)/(3c-2bc+2016)`

`=>M=-1+(3c)/(3c-2bc+2016)-(2bc)/(3c-2bc+2016)-1+(2016)/(3c-2bc+2016)

`=>M=-2+(3c-2bc+2016)/(3c-2bc+2016)`

`=>M=-2+1`

`=>M=-1`

`(2bc-2016)/(3c-2bc+2016)`

`=(-(3c-2bc+2016)+3c)/(3c-2bc+2016)`

`=-1+(3c)/(3c-2bc+2016)`

`(2b)/(3-2b+ab)`

`=(2bc)/(3c-2bc+abc)`

`=(2bc)/(3c-2bc+2016)`

`(4032-3ac)/(3ac-4032+2016a)`

`=(-(3ac-4032+2016a)+2016a)/(3ac-4032+2016a)`

`=-1+(2016a)/(3ac-2abc+2016a)`

`=-1+(2016)/(3c-2bc+2016)`

`=>M=-1+(3c)/(3c-2bc+2016)-(2bc)/(3c-2bc+2016)-1+(2016)/(3c-2bc+2016)`

`=>M=-2+(3c-2bc+2016)/(3c-2bc+2016)`

`=>M=-2+1`

`=>M=-1`

Nãy thiếu latex ạ sorry~~

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

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

\(=\frac{3a+3b+3c+2016}{2015+a}+\frac{3a+3b+3c+2016}{2016+b}+\frac{3a+3b+3c+2016}{2017+c}\)

\(=\left(3a+3b+3c+2016\right)\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\)

\(=4.2016\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\) \(\left(a+b+c=2016\right)\)

\(=8064.\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\)

Vì a ; b ; c dương , áp dụng BĐT phụ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\), ta có :

\(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\ge\frac{9}{2015+2016+2017+a+b+c}=\frac{9}{8064}\)

\(\Rightarrow P+3\ge8064.\frac{9}{8064}=9\) \(\Rightarrow P\ge6\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2015+a=2016+b=2017+c\\a+b+c=2016\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1=c+2\\a+b+c=2016\end{matrix}\right.\)

\(\Leftrightarrow a=673;b=672;c=671\)

Vậy ...

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

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

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

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

Dấu = xảy ra khi ....