\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)

\(\Leftrightarrow\frac{yz\left(x+y+z\right)+xz\left(x+y+z\right)+xy\left(x+y+z\right)-xyz}{xyz\left(x+y+z\right)}=0\)

\(\Leftrightarrow\)\(xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz-xyz=0\)

\(\Leftrightarrow\)\(\left(xyz+y^2z\right)+\left(xyz+x^2z\right)+\left(xz^2+yz^2\right)+\left(xy^2+x^2y\right)=0\)

\(\Leftrightarrow yz\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+xy\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(yz+xz+xy+z^2\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y\\x+z=0\end{cases}}=0\)  hoặc y+z=0

Do đó ta có B=0

Đặt bẫy hả

\(\frac{x+2106}{-3}\)= \(\frac{-12}{x+2106}\)

=> (x+2106)^2 = 36                    (nhân chéo nha bạn )

=> (x+2106)^2=6^2

=> x+2106=6

=> x=-2100

x+2106 . x+2106=-3X12

\(^{X+2106^2}\)=-36

(x+2106)\(^2\)=6\(^2\)hoặc (-6)\(^2\)

x+2106=6 hoặc x+2106=(-6)

x         =6-2106 hoặc x        =(-6)-2106

x= -2100  hoặc x= -2112

12063 x X - 2106 = 37485

12063 x X= 37485+2106

12063 x X= 39591

 Đề sai

tk nhé

12063xX-2106=37485

12063xX=37485+2106

12063xX=39591

sai đề em ơi

ai k mình k lại

bye bye

Ta có

 \(2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=\frac{2016^{2017}+1}{2016^{2017}+1}+\frac{2015}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)

\(2016B=\frac{2016^{2016}+2016}{2016^{2016}+1}=\frac{2016^{2016}+1}{2016^{2016}+1}+\frac{2015}{2016^{2016}+1}=1+\frac{2015}{2016^{2016}+1}\)

Do \(\frac{2015}{2016^{2017}+1}< \frac{2015}{2016^{2016}+1}\Rightarrow2016A< 2016B\Rightarrow A< B.\)

B = \(\frac{2016^{2015}+1}{2016^{2016}+1}\)< A =\(\frac{2016^{2016}+1}{2016^{2017}+1}\)

\(\Leftrightarrow\left(a^{2016}+b^{2016}\right).\left(c^{2016}-d^{2016}\right)=\left(a^{2016}-b^{2016}\right).\left(c^{2016}+d^{2016}\right)\)

\(\Leftrightarrow ac^{2016}-ad^{2016}+bc^{2016}-bd^{2016}=ac^{2016}+ad^{2016}-bc^{2016}-bd^{2016}\)

\(\Leftrightarrow-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}\)

nếu \(-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}=0\)

\(\Rightarrow ad^{2016}-bc^{2016}=0\Rightarrow ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\left(1\right)\)

nếu \(\text{​​}-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}\ne0\Rightarrow ad=-bc\Rightarrow\frac{a}{b}=-\frac{c}{d}\left(2\right)\)

từ (1) và (2) => đpcm

Cho mình hỏi đpcm là gì vậy ? 

Ta có tận cùng của A là 8 . Mà số chính phương không bao giờ có tận cùng là 8 ⇒ A không là số chính phương