cho xyz=1 và x +y+z=\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
tính P = (x\(^{1999}\)-1)(y\(^{2018}\)-1)(z\(^{2019}\)-1)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
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
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: \(\frac{x-y-z}{x}=-1\) \(\Rightarrow\) \(x-y-z=-x\) \(\Rightarrow\) \(y+z=2x\)
\(\frac{-x+y-z}{y}=-1\) \(-x+y-z=-y\) \(x+z=2y\)
\(\frac{-x-y+z}{z}=-1\) \(-x-y+z=-z\) \(x+y=2z\)
\(A=\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\)
\(=\frac{x+y}{x}.\frac{y+z}{y}.\frac{x+z}{z}\)
\(=\frac{2z}{x}.\frac{2x}{y}.\frac{2y}{z}=8\)
Lời giải:
Vì $xyz=1$ nên:
\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xyz}{x}+\frac{xyz}{y}+\frac{xyz}{z}=xy+yz+xz\)
\(\Leftrightarrow x+y+z-xy-yz-xz=0\)
\(\Leftrightarrow 1+x+y+z-xy-yz-xz-1=0\)
\(\Leftrightarrow xyz+x+y+z-xy-yz-xz-1=0\)
\(\Leftrightarrow xy(z-1)+(x+y-yz-xz)+(z-1)=0\)
\(\Leftrightarrow xy(z-1)-x(z-1)-y(z-1)+(z-1)=0\)
\(\Leftrightarrow (z-1)(xy-x-y+1)=0\)
\(\Leftrightarrow (z-1)(x-1)(y-1)=0\)
Do đó:
\(P=(x^{1999}-1)(y^{2018}-1)(z^{2019}-1)\)
\(=(x-1)(x^{1998}+x^{1997}+...+1)(y-1)(y^{2017}+...+1)(z-1)(z^{2018}+....+1)\)
\(=(x-1)(y-1)(z-1).A=0.A=0\)