- 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~~

Dặt x=a, y=2b,z=3c

Khi đó

\(P=\frac{yz}{\sqrt{x+yz}}+\frac{xz}{\sqrt{y+xz}}+\frac{xy}{\sqrt{z+xy}}\)và x+y+z=1

Ta có \(\frac{yz}{\sqrt{x+yz}}=\frac{yz}{\sqrt{x\left(x+y+z\right)+yz}}=\frac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\frac{1}{2}yz\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)

=> \(P\le\frac{1}{2}\left(\frac{xz}{x+y}+\frac{yz}{x+y}\right)+\frac{1}{2}\left(\frac{xy}{y+z}+\frac{xz}{y+z}\right)+...=\frac{1}{2}\left(x+y+z\right)\)

                                                                                                                     \(=\frac{1}{2}\)

Vậy \(MaxP=\frac{1}{2}\)khi x=y=z=1/3 hay \(\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{1}{6}\\c=\frac{1}{9}\end{cases}}\)

\(2016-4032:m-4032:2015\)

Bạn thêm điều kiện x,y,z lớn hơn 0 nhé :)

Từ giả thiết ta suy ra : \(a^2=b+4032\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4032\)

\(\Rightarrow xy+yz+zx=2016\)thay vào :

\(x\sqrt{\frac{\left(2016+y^2\right)\left(2016+z^2\right)}{2016+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}\)

\(=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+y\right)\left(z+x\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=x\left|y+z\right|=xy+xz\)vì x,y,z > 0

Tương tự : \(y\sqrt{\frac{\left(2016+z^2\right)\left(2016+x^2\right)}{2016+y^2}}=xy+zy\)

\(z\sqrt{\frac{\left(2016+x^2\right)\left(2016+y^2\right)}{2016+z^2}}=zx+zy\)

Suy ra \(P=2\left(xy+yz+zx\right)=2.2016=4032\)