Ta có :

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

\(\Rightarrow z\left(x+y\right)=x\left(y+z\right)=y\left(z+x\right)\)

Từ \(z\left(x+y\right)=x\left(y+z\right)\Leftrightarrow xz+yz=xy+xz\Leftrightarrow yz=xy\Rightarrow x=z\) (1)

Từ \(x\left(y+z\right)=y\left(x+z\right)\Leftrightarrow xy+xz=xy+yz\Leftrightarrow xz=yz\Rightarrow x=y\) (2)

Từ \(z\left(x+y\right)=y\left(z+x\right)\Leftrightarrow xz+yz=yz+xy\Leftrightarrow xz=xy\Rightarrow z=y\) (3)

Từ (1) ; (2) ; (3) \(\Rightarrow x=y=z\) (đpcm)

Ta có \(xy+xz+yz=xyz\left(x+y+z\right)\)

\(\Leftrightarrow x+y+z=\frac{xy+xz+yz}{xyz}\left(1\right)\)

Ta lại có \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}=\frac{x^2-yz-y^2+xz}{x\left(1-yz\right)-y\left(1-xz\right)}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)

Vậy ta có đpcm

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\)

\(\Rightarrow\left(x^2-yz\right)y\left(1-yz\right)=\left(y^2-xz\right)x\left(1-yz\right)\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2=xy^2-x^2z-xy^3z+x^2yz^2\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+x^2z+xy^3z-x^2yz^2=0\)

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

\(\Rightarrow\left(x-y\right)\left[xy-xyz\left(x+y+z\right)+xz+yz\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy+yz+zx=0\end{cases}}\)

Mà \(x\ne y\) nên \(xy+xz+yz-xyz\left(x+y+z\right)=0\)

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

Đpcm

Từ gt ta có : (x² - yz)y(1 - yz) = (y² - xz)x(1 - yz)

=> 0 = VT - VP = (x²y - x³yz - y²z - xy²z²) - (xy² - xy³z  - x²z - x²yz²) = xy(x - y) - xyz(x² - y²) + z(x² - y²) + xyz²(y - x)

= (x - y)[xy - xyz(x + y) + z(x + y) - xyz²] = (x - y)(xy + yz + xz - xyz(x + y + z)]

Vì\(x\ne y\Rightarrow x-y\ne0\) nên xy + yz + xz - xyz(x + y + z) = 0 => xy + yz + xz = xyz(x + y + z)

Bạn ko hiểu chỗ nào thì hỏi mình nhé!

ĐƠN GIẢN

x²+y²+z²=xy+yz+zx

<=>2(x²+y²+z²)=2(xy+yz+zx)

<=>2x²+2y²+2z²=2xy+2yz+2zx

<=>2x²+2y²+2z²-2xy-2yz-2zx=0

<=>(x²-2xy+y²)+(y²-2yz+z²)+(z²-2zx+x²)=0

<=>(x-y)²+(y-z)²+(z-x)²=0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow x=y=z}\)(đpcm)

SORY I'M I GRADE 6

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mày hỏi vả bài kiểm tra à thằng điên 

Áp dụng BĐT AM-GM ta có:

\(\frac{\left(y+z\right)\sqrt{yz}}{x}\ge\frac{2\sqrt{yz}\cdot\sqrt{yz}}{x}=\frac{2\sqrt{\left(yz\right)^2}}{x}=\frac{2yz}{x}\)

Tương tự cho 2 BĐT còn lại ta cũng có

\(\frac{\left(x+y\right)\sqrt{xy}}{z}\ge\frac{2xy}{z};\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xz}{y}\)

\(\Leftrightarrow\frac{\left(y+z\right)\sqrt{yz}}{x}+\frac{\left(x+y\right)\sqrt{xy}}{z}+\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\)

Cần chứng minh \(\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\ge2\left(x+y+z\right)\)

\(\Leftrightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge x+y+z\)

Áp dụng BĐT AM-GM:

\(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}\cdot\frac{yz}{x}}=2\sqrt{y^2}=2y\)

Tương tự rồi cộng theo vế ta có ĐPCM

Khi \(x=y=z\)

Ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2.\)

<=>        \(x^2+2xy+y^2+2xz+2yz+z^2-x^2-y^2-z^2=0\)

<=>         \(2xy+2xz+2yz=0\)

<=>          \(2.\left(xy+xz+yz\right)=0\)

<=>           \(xy+xz+yz=0\)

Vậy_

Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2\)

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

\(\Leftrightarrow2\left(xy+xz+yz\right)=0\)

\(xy+xz+yz=0\left(đpcm\right)\)