Sửa đề : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2019}\)

Thay \(2019=x+y+z\)ta có :

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

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

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

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

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

\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\left(x+y\right)\)

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

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

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

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

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

( mình chỉ xét 1 t/h, các t/h còn lại hoàn toàn tương tự )

TH1 : \(x+y=0\)

\(\Leftrightarrow x=-y\)(1)

Thay (1) vào A ta có :

\(A=\frac{1}{-y^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}\)

\(A=\frac{1}{z^{2019}}\)

Mặt khác : \(x+y+z=2019\)

Thay (1) vào đẳng thức trên ta được : \(-y+y+z=2019\)

\(\Leftrightarrow z=2019\)

Thay z vào A ta được : \(A=\frac{1}{2019^{2019}}\)

sửa đền nha:^{\(\frac{1}{x}\)}+\(\frac{1}{y}\)+\(\frac{1}{z}\)=\(\frac{1}{2019}\)

nguyệt dạ hình như là mai linh... t đang tìm bài này thì thấy lp mk ... thành ra trong lp hỏi nhau :)))

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

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

\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)

\(\Rightarrow A=0\)

Sai đề không

Sai đề thật, cái biểu thức trên 100% lớn hơn hoặc = 9, lấy đâu ra =1

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)=> \(\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)

=> (x+y+z)(xy+yz+zx) = xyz

=> \(x^2y+xy^2+y^2z+yz^2+zx^2+z^2x+2xyz=0\)

=> (x+y)(y+z)(z+x) = 0

=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

TH1: x = -y

=> \(\frac{1}{x^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{\left(-y\right)^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{z^{2019}}\)

=> \(\frac{1}{x^{2019}+y^{2019}+z^{2019}}=\frac{1}{\left(-y\right)^{2019}+y^{2019}+z^{2019}}=\frac{1}{z^{2019}}\)

=> ĐPCM

Tương tự với TH2 và TH3

Đk: $x\geq \frac{1}{2}$

Pt $\Leftrightarrow 4x^2+3x-7=4(\sqrt{x^3+3x^2}-2)+2(\sqrt{2x-1}-1)$

$\Leftrightarrow +4\frac{(x-1)(x+2)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-(x-1)(4x+7)=0$

$\Leftrightarrow (x-1)[\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-(4x+7)]=0$

$\Leftrightarrow x=1\vee \frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0$ $(*)$

Xét hàm số $f(x)=\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\in [\frac{1}{2};+\infty )$ thì $f(x)>0,\forall x\in [\frac{1}{2};+\infty )$

$\Rightarrow $ Pt $(*)$ vô nghiệm

Ta có : \(A=\frac{2019}{x+xy+1}+\frac{2019}{y+yz+1}+\frac{2019}{z+zx+1}=2019\left(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+xyz+z}+\frac{xz}{xyz+xyz^2+xz}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+z+1}+\frac{xz}{1+z+xz}+\frac{1}{z+zx+1}\right)\)(vì xyz = 1)

\(=2019\left(\frac{z+xz+1}{xz+z+1}\right)=2019\)

Vậy A = 2019