\(S=\frac{yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

+ \(yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)\)

\(=yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left[\left(y-z\right)+\left(x-y\right)\right]\)

\(+xy\left(z+1\right)\left(x-y\right)\)

\(=\left(y-z\right)\left[yz\left(x+1\right)-zx\left(y+1\right)\right]+\left(x-y\right)\left[xy\left(z+1\right)-zx\left(y+1\right)\right]\)

\(=\left(y-z\right)\left[z\left(y-x\right)\right]+\left(x-y\right)\cdot x\cdot\left(y-z\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

\(\Rightarrow S=\frac{1}{xyz}\)

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)

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

\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\Leftrightarrow\left(x+y\right)\left(\frac{zx+z^2+zy+xy}{xyz\left(x+y+z\right)}\right)=0\)

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

\(\Rightarrow\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=0\).

Vậy  \(M=\frac{3}{4}+\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=\frac{3}{4}+0=\frac{3}{4}\)

thank Gia Hy

mong mọi người nhanh giúp

Thế câu một các cậu làm được chưa

Ta có : \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)

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

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{y+z}{x}=\frac{x+z}{y}=\frac{x+y}{z}=\frac{y+z+x+z+x+y}{x+y+z}=2\)

+) \(\frac{y+z}{x}=2\)

=> y+z=2x

+) \(\frac{x+z}{y}=2\)

=>x+z=2y

+)\(\frac{x+y}{z}=2\)

=> x+y=2z 

Mà B= ( 1+x/y)(1+y/z) (1+z/x)

      B= \(\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}\)

      B= \(\frac{2z.2x.2y}{xyz}\)

      B= 8

~ Chúc bạn học tốt ~

Tích và kết bạn với mình nha!

Ta có: \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)

Lại có:

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

\(\Leftrightarrow\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)

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

(+) Xét x + y + z = 0\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}}\)

Thay vào ta có: \(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=\frac{-xyz}{xyz}=-1\)

(+) Xét x + y + z \(\ne\) 0

Tương tự như trên ta có: \(\hept{\begin{cases}x+y=2z\\y+z=2x\\z+x=2y\end{cases}}\)

Thay vào ta có: \(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)

Vậy \(\hept{\begin{cases}B=-1\Leftrightarrow x+y+z=0\\B=8\Leftrightarrow x+y=y+z=z+x\Leftrightarrow x=y=z\end{cases}}\)

\(A\le\left|A\right|=\dfrac{\left|xy+yz+xz\right|}{\left|xyz\right|}\)

Áp dụng: \(\left|a+b+c\right|\le\left|a\right|+\left|b\right|+\left|c\right|\)

\(\left|A\right|\le\dfrac{\left|xy\right|+\left|yz\right|+\left|xz\right|}{\left|xyz\right|}=\dfrac{1}{\left|x\right|}+\dfrac{1}{\left|y\right|}+\dfrac{1}{\left|z\right|}\)

\(\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)

Ta có đpcm. Dấu "=" khi \(x=y=z=3\)

Thêm 1 hướng suy nghĩ khác
Ta có: \(\left|x\right|\ge3;\left|y\right|\ge3;\left|z\right|\ge3\)

\(\Rightarrow0< \dfrac{1}{\left|x\right|}\le\dfrac{1}{3};0< \dfrac{1}{\left|y\right|}\le\dfrac{1}{3};0< \dfrac{1}{\left|z\right|}\le\dfrac{1}{3}\)

Ta có:

\(A=\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{1}{\left|x\right|}+\dfrac{1}{\left|y\right|}+\dfrac{1}{\left|z\right|}\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)

vì x + 2 = y + 1 = z + 3 => x = y - 1 = z + 1 ; y = x + 1 = z + 2; z = x + 1 = y - 2  và z < x < y

ta có (x-1/3).(y-1/2).(z-5)=0 => ta có 3 TH

TH₁ z - 5 = 0 => z = 5 ; y = 7 ; x = 4

TH₂ x - 1/3 = 0 => x = 1/3 ; y = 4/3 ; z = -2/3

TH₃ y - 1/2 = 0 => y = 1/2 ; x = -1/2 ; z = -3/2

nhớ cho mik nha 

Ta có:

\(\left(x-\frac{1}{2}\right).\left(y-\frac{1}{2}\right).\left(z-5\right)=0\)

\(\Rightarrow x-\frac{1}{2}=0;y-\frac{1}{2}=0\)hoặc \(z-5=0\)

Với \(x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)\(\Rightarrow\)\(x+2=\frac{1}{3}+2=\frac{7}{3}=y+1=z+3\)\(\Rightarrow y=...;z=...\)

Với \(y-\frac{1}{2}=0\Rightarrow y=\frac{1}{2}\)\(\Rightarrow....\)

Với \(z-5=0\)\(\Rightarrow.....\)

B tự làm nốt nhé

nhớ k nha