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TH1: \(x+y+z+t=0\)
\(P=\left(1+\dfrac{x+y}{z+t}\right)^{2023}+\left(1+\dfrac{y+z}{x+t}\right)^{2023}+\left(1+\dfrac{z+t}{x+y}\right)^{2023}+\left(1+\dfrac{t+x}{y+z}\right)^{2023}\)
\(=\left(\dfrac{x+y+z+t}{z+t}\right)^{2023}+\left(\dfrac{x+y+z+t}{x+t}\right)^{2023}+\left(\dfrac{x+y+z+t}{x+y}\right)^{2023}+\left(\dfrac{x+y+z+t}{y+z}\right)^{2023}\)
\(=0+0+0+0=0\) là số nguyên (thỏa mãn)
TH2: \(x+y+z+t\ne0\), áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2023x+y+z+t}=\dfrac{y}{x+2023y+z+t}=\dfrac{z}{x+y+2023z+t}+\dfrac{t}{x+y+z+2023t}\)
\(=\dfrac{x+y+z+t}{\left(2023x+y+z+t\right)+\left(x+2023y+z+t\right)+\left(x+y+2023z+t\right)+\left(x+y+z+2023t\right)}\)
\(=\dfrac{x+y+z+t}{2026\left(x+y+z+t\right)}=\dfrac{1}{2026}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2023x+y+z+t}=\dfrac{1}{2026}\\\dfrac{y}{x+2023y+z+t}=\dfrac{1}{2026}\\\dfrac{z}{x+y+2023z+t}=\dfrac{1}{2026}\\\dfrac{t}{x+y+z+2023t}=\dfrac{1}{2026}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2026x=2023x+y+z+t\\2026y=x+2023y+z+t\\2026z=x+y+2023z+t\\2026t=x+y+z+2023t\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}4x=x+y+z+t\\4y=x+y+z+t\\4z=x+y+z+t\\4t=x+y+z+t\end{matrix}\right.\)
\(\Rightarrow4x=4y=4z=4t\) (vì đều bằng \(x+y+z+t\))
\(\Rightarrow x=y=z=t\)
Do đó:
\(P=\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}\)
\(=2^{2023}+2^{2023}+2^{2023}+2^{2023}\)
\(=4.2^{2023}=2^{2025}\in Z\)
Em kiểm tra lại đề, 2 ngoặc cuối bị giống nhau, chắc em ghi nhầm
\(x+\frac{1}{y}=y+\frac{1}{z}\Rightarrow x-y=\frac{1}{z}-\frac{1}{y}=\frac{z-y}{zy}\)
\(y+\frac{1}{z}=z+\frac{1}{x}\Rightarrow y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{xz}\)
\(z+\frac{1}{x}=x+\frac{1}{y}\Rightarrow z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}\)
\(\Rightarrow\left(x-y\right)\left(y-z\right)\left(z-x\right)=\frac{y-z}{zy}\cdot\frac{z-x}{zx}\cdot\frac{x-y}{xy}\)
\(\Rightarrow\left(x-y\right)\left(y-z\right)\left(z-x\right)=\frac{\left(y-z\right)\left(z-x\right)\left(x-y\right)}{x^2y^2z^2}\)
\(\Rightarrow x^2y^2z^2\left(x-y\right)\left(y-z\right)\left(z-x\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
\(\Rightarrow\left(x^2y^2z^2-1\right)\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2y^2z^2-1=0\\\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2y^2z^2=1\\x=y=z\end{cases}}\)
x+1/y=y+1/z => x-y=1/z-1/y=(y-z)/yz
Tương tự y-z=(z-x)/zx ; z-x=(x-y)/xy
Nhân theo vế các đẳng thức trên ta đc:
(x-y)(y-z)(z-x)=(x-y)(y-z)(z-x)/x2y2z2
=>(x-y)(y-z)(z-x)x2y2z2-(x-y)(y-z)(z-x)=0
=>(x-y)(y-z)(z-x)(x2y2z2-1)=0
=>x-y=0 hoặc y-z=0 hoặc z-x=0 hoặc x2y2z2-1=0
=>x=y=z hoặc x2y2z2=1(đfcm)
Ta có \(\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\). Áp dụng tính chất dãy tỉ số bằng nhau
=>\(\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}=\frac{y+x+y+x}{x-z+z+y}=\frac{2\left(x+y\right)}{x+y}=2\)
=>\(\frac{x}{y}=2=>x=2y\)
Có \(\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\left(x\ne y\ne z;x,y,z>0\right)\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}=\frac{y+x+y+x}{x-z+z+y}=\frac{2\left(x+y\right)}{x+y}=2\)
\(\Rightarrow\frac{x}{y}=2\Rightarrow x=2y\left(đpcm\right)\)
Ta có: \(x+\frac{1}{y}=y+\frac{1}{z}\Rightarrow x-y=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{yz}\)(1)
\(y+\frac{1}{z}=z+\frac{1}{x}\Rightarrow y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{zx}\)(2)
\(z+\frac{1}{x}=x+\frac{1}{y}\Rightarrow z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}\)(3)
Nhân vế theo vế ba đẳng thức (1), (2), (3), ta được: \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{x^2y^2z^2}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\left(^∗\right)\\x^2y^2z^2=1\end{cases}}\)
Từ (*) ta giả sử x - y = 0 thì x = y khi đó \(\frac{1}{y}=\frac{1}{z}\Rightarrow y=z\)suy ra x = y = z. Tương tự đối với y - z = 0; z - x = 0
Vậy x = y = z hoặc x2y2z2 = 1
Ta có:\(x:y:z=1:2:3\Rightarrow x=\frac{y}{2}=\frac{z}{3}\).Đặt \(x=\frac{y}{2}=\frac{z}{3}=k\)
\(\Rightarrow\hept{\begin{cases}x=k\\y=2k\\z=3k\end{cases}}\)\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=\left(k+2k+3k\right)\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)
\(=6k.\left(\frac{1}{k}+\frac{2}{k}+\frac{3}{k}\right)=6k.\frac{6}{k}=36\)
\(\Rightarrowđpcm\)
Ta có :\(\frac{1}{x}=\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
=> \(\frac{1}{x}=\frac{y+z}{2yz}\)
=> 2yz = x(y + z)
=> 2yz - xy - xz = 0
=> (yz - xy) + (yz - xz) = 0
=> y(z - x) + z(y- x) = 0
=> y(z - x) = -z(y - x)
=> -y(x - z) = -z(y - x)
=> \(\frac{-z}{-y}=\frac{x-z}{y-x}\Leftrightarrow\frac{z}{y}=\frac{x-z}{y-x}\)