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Theo bài ra, ta có:
\(P=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)
\(=\frac{x}{xy+x+1}+\frac{xy}{x\left(yz+y+1\right)}+\frac{z}{xz+z+xyz}\)
\(=\frac{x}{xy+x+1}+\frac{xy}{xyz+xy+x}+\frac{z}{z\left(x+1+xy\right)}\)
\(=\frac{x}{xy+x+1}+\frac{xy}{xy+x+1}+\frac{1}{xy+x+1}\)
\(=\frac{x+xy+1}{xy+x+1}\)
\(=1\)
Vậy P = 1
Ta có: P = \(\dfrac{x}{xy+x+1}\)+\(\dfrac{y}{yz+y+1}\)+\(\dfrac{z}{xz+z+1}\)
=\(\dfrac{x}{xy+x+1}\)+\(\dfrac{xy}{xyz+xy+x}\)+\(\dfrac{xyz}{x^2yz+xyz+xy}\)
=\(\dfrac{x}{xy+x+1}\)+\(\dfrac{xy}{xy+x+1}\)+\(\dfrac{1}{xy+x+1}\)(vì xyz=1)
=\(\dfrac{x+xy+1}{xy+x+1}\)
=1
Vậy P = 1
Lời giải:
Ta có: \(x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)
Vì \((x-y)^2; (y-z)^2;(z-x)^2\geq 0\), do đó để tổng của chúng bằng $0$ thì:
\((x-y)^2=(y-z)^2=(z-x)^2=0\Rightarrow x=y=z\)
\(\Rightarrow 3x^{2017}=3y^{2017}=3z^{2017}=x^{2017}+y^{2017}+z^{2017}=9\)
\(\Rightarrow x=y=z=\sqrt[2017]{3}\)
\(\Rightarrow \left(\frac{2017x+2018y-4023z}{3}\right)^{2017}=\left(\frac{12x}{3}\right)^{2017}=(4x)^{2017}=3.4^{2017}\)
Có \(\frac{2020x}{xy+2020x+2020}=\frac{2020}{y+2020+yz}\) (1)và \(\frac{z}{xz+z+1}=\frac{yz}{2020+yz+y}\)(2)
coog (1) và (2) và y/yz+y+2020 có
ĐPCM
THAY 2018 = xyz vào biểu thức
\(\frac{xyzx}{xy+xyzx+xyz}\) + \(\frac{y}{yz+y+xyz}\)+ \(\frac{z}{xz+z+1}\)
= \(\frac{xz}{1+xz+z}\)+ \(\frac{1}{z+1+xz}\)+ \(\frac{z}{xz+z+1}\)= \(\frac{xz+z+1}{xz+z+1}\)=\(1\)
Đặt \(A=\frac{2018x}{xy+2018x+2018}+\frac{y}{yzz+y+2018}+\frac{z}{xz+z+1}\)
Thay \(xyz=2018\)vào A ta được
\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{1}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz+1+z}{xz+z+1}=1\)
\(\frac{2011x}{xy+2011x+2011}+\frac{y}{yz+y+2011}+\frac{z}{zx+z+1}\)
\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{zx+z+1}\)
\(=\frac{x^2yz}{xy.\left(xz+z+1\right)}+\frac{y}{y.\left(xz+z+1\right)}+\frac{z}{zx+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{zx+z+1}\)
\(=\frac{xz+1+z}{xz+1+z}\)
\(=1\)
đpcm
Tại sao lại có nhìu đứa rảnh háng đi trả lời câu này nhỉ ?
thay xyz=2017 vaf 2017=xyz a đc :
\(\frac{xyz.x}{xy+xyz.x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)=\(\frac{xyz.x}{xy.\left(xz+z+1\right)}+\frac{y}{y.\left(xz+z+1\right)}+\frac{z}{xz+z+1}\)
=\(\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=\frac{xz+z+1}{xz+z+1}=1\)