cho x,y,z khac 0 va\(\dfrac{x+3y-z}{z}\)= \(\dfrac{y+3z-x}{x}\)=\(\dfrac{z+3x-y}{y}\)
Tính P = \(\left(\dfrac{x}{y}+3\right)\)\(\left(\dfrac{y}{z}+3\right)\)\(\left(\dfrac{z}{x}+3\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
20 tháng 11 2021
Xét \(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\\z+x=-y\\x+y=-z\end{matrix}\right.\)
\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)
Xét \(x+y+z\ne0\) thì ta có:
\(\dfrac{x}{y+z+3x}=\dfrac{y}{z+x+3y}=\dfrac{z}{x+y+3z}=\dfrac{x+y+z}{5x+5y+5z}=\dfrac{x+y+z}{5\left(x+y+z\right)}=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)
\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)
Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)
20 tháng 11 2021
Nếu bị lỗi thì bạn có thể xem đây nhé:
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(P=\dfrac{1}{3x\left(y+z\right)+x+y+z}+\dfrac{1}{3y\left(z+x\right)+x+y+z}+\dfrac{1}{3z\left(x+y\right)+x+y+z}\)
\(P\le\dfrac{1}{3x\left(y+z\right)+3\sqrt[3]{xyz}}+\dfrac{1}{3y\left(z+x\right)+3\sqrt[3]{xyz}}+\dfrac{1}{3z\left(x+y\right)+3\sqrt[3]{xyz}}\)
\(P\le\dfrac{1}{3x\left(y+z\right)+3}+\dfrac{1}{3y\left(z+x\right)+3}+\dfrac{1}{3z\left(x+y\right)+3}\)
Đặt \(\left(x;y;z\right)=\left(a^3;b^3;c^3\right)\Rightarrow abc=1\)
\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{1}{a^3\left(b^3+c^3\right)+1}+\dfrac{1}{b^3\left(c^3+a^3\right)+1}+\dfrac{1}{c^3\left(a^3+b^3\right)+1}\right)\)
\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{1}{a^3bc\left(b+c\right)+1}+\dfrac{1}{b^3ac\left(a+c\right)+1}+\dfrac{1}{c^3ab\left(a+b\right)+1}\right)\)
\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{bc}{a\left(b+c\right)+bc}+\dfrac{ac}{b\left(a+c\right)+ac}+\dfrac{ab}{c\left(a+b\right)+ab}\right)=\dfrac{1}{3}\)
\(P_{max}=\dfrac{1}{3}\) khi \(a=b=c=1\) hay \(x=y=z=1\)
2 tháng 1 2023
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
mong mn help
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{x+3y-z+y+3z-x+z+3x-y}{x+y+z}=\dfrac{3(x+y+z)-(x+y+z)}{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}=2\)
\(\Rightarrow x=y=z=0\)
\(\Rightarrow \) P không xác định. (?)