cho xyz khác 0 và \(\frac{x-y-z}{x}=\frac{-x+y-z}{y}=\frac{-x-y+z}{z}\) tính \(A=(1+\frac{y}{x})(1+\frac{z}{y})(1+\frac{x}{z})\)
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\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
\(A=\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{y+z}{z}\)
Do \(x-y-z=0\)
\(\Rightarrow x-z=y;y-x=-z;y+z=x\)
Khi đó \(A=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)
Vậy A=-1
\(\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{xyz+yz+y}\)
\(=\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{1+yz+y}\)
\(=\frac{1}{xy+x+1}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz}{xy\cdot yz+xyz+yz}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz}{yz+y+1}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz+y+1}{yz+y+1}\)
\(=1\)
Ta có: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\right)\)
\(\left(\sqrt{3}\right)^2=P+\frac{2\left(z+y+x\right)}{xyz}\)
Mà x+y+z=xyz
=> P+2=3=>P=1
Vậy P=1
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
+ Nếu x + y + z = 0 => x + y = -z; y + z = -x; x + z = -y
A = (1 + y/x)(1 + z/y)(1 + x/z)
A = (x+y)/x . (y+z)/y . (x+z)/z
A = -z/x . (-x)/y . (-y)/z = -1
+ Nếu x + y + z khác 0
x-y-z/x = -x+y-z/y = -x-y+z/z
<=> 1 - (y+z)/x = 1 - (x+z)/y = 1 - (x+y)/z
<=> y+z/x = x+z/y = x+y/z
Áp dụng t/c của dãy tỉ số = nhau ta có:
y+z/x = x+z/y = x+y/z = 2(x+y+z)/x+y+z = 2
A = (x+y)/x . (y+z)/y . (x+z)/z = 8
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow xy+yz+xz=0\)
CM : \(x^3y^3+y^3z^3+x^3z^3=3x^2y^2z^2\)
CM: \(x+y+z=0\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Rightarrow\frac{x^6+y^6+z^6}{x^3+y^3+z^3}=\frac{\left(x^3+y^3+z^3\right)^2-2\left(x^3y^3+x^3z^3+y^3z^3\right)}{3xyz}=\frac{3x^2y^2z^2}{xyz}=xyz\)
\(\frac{y+z}{x}=\frac{x+z}{y}=\frac{x+y}{z}\Rightarrow k=2\Rightarrow x=y=z=1\)
A=6
\(\frac{x-y-z}{x}=1-\frac{y+z}{x}\) tương tự con khác
=> x=y=z
=> A=6
\(x+y+z=xyz\Rightarrow\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}=1\)
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=1,63205^2\Rightarrow C+2\left(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\right)=1.63205^2\)
\(\Rightarrow C+2=1.63205^2\Rightarrow C=1.63205^2-2\)
\(\frac{x-y-z}{x}=\frac{y-x-z}{y}=\frac{z-x-y}{z}=\frac{x-y-z+y-x-z+z-x-y}{x+y+z}=\frac{-x-y-z}{x+y+z}=-1\)
\(\rightarrow\begin{cases}x-y-z=-x\\y-x-z=-y\\z-x-y=-z\end{cases}\)
\(\leftrightarrow\begin{cases}y+z=2x\\z+x=2y\\x+y=2z\end{cases}\)
\(A=\frac{x+y}{z}.\frac{y+z}{x}.\frac{z+x}{y}=8\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-y-z}{x}=\frac{-x+y-z}{y}=\frac{-x-y+z}{z}=\frac{x-y-z-x+y-z-x-y+z}{x+y+z}\)\(=\frac{-\left(x+y+z\right)}{x+y+z}\)
Nếu \(x+y+z=0\)thì \(\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}}\)
\(A=\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\)
\(=\frac{x+y}{x}.\frac{y+z}{y}.\frac{z+x}{z}\)
\(=\frac{-z}{x}.\frac{-x}{y}.\frac{-y}{z}=-1\)
Nếu \(x+y+z\ne0\)thì \(\frac{x-y-z}{x}=\frac{-x+y-z}{y}=\frac{-x-y+z}{z}=-1\)
suy ra: \(\frac{x-y-z}{x}=-1\) \(\Rightarrow\) \(x-y-z=-x\) \(\Rightarrow\) \(y+z=2x\)
\(\frac{-x+y-z}{y}=-1\) \(-x+y-z=-y\) \(x+z=2y\)
\(\frac{-x-y+z}{z}=-1\) \(-x-y+z=-z\) \(x+y=2z\)
\(A=\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\)
\(=\frac{x+y}{x}.\frac{y+z}{y}.\frac{x+z}{z}\)
\(=\frac{2z}{x}.\frac{2x}{y}.\frac{2y}{z}=8\)