Ta có: \(\dfrac{x-y}{z-y}=-10\)

nên \(z-y=\dfrac{x-y}{-10}\)

hay \(y-z=\dfrac{x-y}{10}=\dfrac{1}{10}\left(x-y\right)\)

Ta có: \(\dfrac{x-y}{z-y}=-10\)

\(\Leftrightarrow\dfrac{x-y}{-10}=\dfrac{z-y}{1}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-y}{-10}=\dfrac{z-y}{1}=\dfrac{x-y-z+y}{-10-1}=\dfrac{x-z}{-11}\)

Do đó: \(\dfrac{x-y}{-10}=\dfrac{x-z}{-11}\)

\(\Leftrightarrow x-z=\dfrac{11\left(x-y\right)}{10}=\dfrac{11}{10}\left(x-y\right)\)

\(\Leftrightarrow\dfrac{x-z}{y-z}=\dfrac{11}{10}\left(x-y\right):\dfrac{1}{10}\left(x-y\right)=\dfrac{11}{10}\cdot\dfrac{10}{1}=11\)

Câu trả lời là thiếu dự kiện

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\) (*)

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

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

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

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

Thay (*) vào,ta có : \(A=x.\left(\frac{-1}{x}\right)+y.\left(-\frac{1}{y}\right)+z.\left(-\frac{1}{z}\right)=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)

thay z = -(x+y) , y = -(z+x),... vao

=> Duoc bieu thuc trong do co 1/xy + 1/yz + 1/zx = (x+y+z)/xyz = 0

\(x+y+z=0\Rightarrow x+y=-z\)

\(\Rightarrow\left(x+y\right)^2=\left(-z\right)^2\Rightarrow x^2+2xy+y^2=z^2\Rightarrow x^2+y^2-z^2=-2xy\)

Tương tự: \(y^2+z^2-x^2=-2yz,x^2+z^2-y^2=-2xz\)

\(\frac{1}{y^2+z^2-x^2}+\frac{1}{x^2+y^2-z^2}+\frac{1}{x^2+z^2-y^2}\)

\(=\frac{1}{-2yz}+\frac{1}{-2xy}+\frac{1}{-2xz}=\frac{x+y+z}{-2xyz}=0\)

Ta có : \(x+y+z=0\Rightarrow x+y=-z\)

\(\Rightarrow\left(x+y\right)^2=z^2\Rightarrow x^2+y^2+2xy=z^2\)

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

Tương tự ta có : \(y^2+z^2=x^2-2yz\)

\(x^2+z^2=y^2-2xz\)

Thay vào biểu thức ta có :

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

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

\(=-\frac{x^2}{2yz}-\frac{y^2}{2xz}-\frac{z^2}{2xy}\)

\(=\frac{-x^3-y^3-z^3}{2xyz}=-\frac{x^3+y^3+z^3}{2xyz}\)

\(=\frac{3xyz}{2xyz}=-\frac{3}{2}\)

Chỗ \(x^3+y^3+z^3=3xyz\)là do \(x+y+z=0\)nhé, bạn cần chứng minh không ?

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

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

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

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

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

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

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

\(=y.\frac{-1}{y}+x.\frac{-1}{x}+z.\frac{-1}{z}=-1-1-1=-3\)

Vậy nên A = -3

\(a^2-2b+6b+b^2=-10\)

\(\Leftrightarrow a^2-2a+6b+b^2+10=0\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+6b+9\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\left(1\right)\)

Vì \(\hept{\begin{cases}\left(a-1\right)^2\ge0\forall a\\\left(b+3\right)^2\ge0\forall b\end{cases}\Leftrightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+3\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=1\\b=-3\end{cases}}}\)

\(L=\frac{x+y}{z}+1+\frac{y+z}{x}+1+\frac{x+z}{y}+1-3\)

\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=0-3=-3\)