(1 điểm) Cho $ {x}, {y}, {z} \neq 0$ và $\dfrac{x+2 y-z}{z}=\dfrac{y+2 z-x}{x}=\dfrac{z+2 x-y}{y}$.
Tính $P=\left(\dfrac{x}{y}+2\right)\left(\dfrac{y}{z}+2\right)\left(\dfrac{z}{x}+2\right)$.
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Những câu hỏi liên quan
26 tháng 11 2022
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
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:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
AH
Akai Haruma
Giáo viên
10 tháng 1 2022
Lời giải:
\(A=\left(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}\right)\left(\frac{1}{y-z}+\frac{1}{z-x}+\frac{1}{x-y}\right)-\frac{x}{(y-z)(z-x)}-\frac{x}{(y-z)(x-y)}-\frac{y}{(z-x)(x-y)}-\frac{y}{(z-x)(y-z)}-\frac{z}{(x-y)(y-z)}-\frac{z}{(x-y)(z-x)}\)
\(=0-\frac{x(x-y)+x(z-x)+y(y-z)+y(x-y)+z(z-x)+z(y-z)}{(x-y)(y-z)(z-x)}\)
\(=0-\frac{x^2+xz+y^2+xy+z^2+zy-(xy+x^2+yz+y^2+zx+z^2)}{(x-y)(y-z)(z-x)}=0-\frac{0}{(x-y)(y-z)(z-x)}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 12 2018
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
28 tháng 11 2022
a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)
21 tháng 11 2017
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
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:
\(P=\left(\dfrac{x+2y}{y}\right)\left(\dfrac{y+2z}{z}\right)\left(\dfrac{z+2x}{x}\right)\)
Ta có
\(\dfrac{x+2y-z}{z}=\dfrac{y+2z-x}{x}=\dfrac{z+2x-y}{y}=\)
\(=\dfrac{x+2y-z+y+2z-x+z+2x-y}{x+y+z}=\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\dfrac{x+2y}{z}-1=\dfrac{y+2x}{x}-1=\dfrac{z+2x}{y}-1=2\)
\(\Rightarrow\dfrac{x+2y}{z}=\dfrac{y+2x}{x}=\dfrac{z+2x}{y}=3\)
\(\Rightarrow P=3.3.3=27\)