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\(D=\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
Vì \(xyz=2017\)
\(D=\dfrac{xy\left(xz\right)}{xy\left(1+xz+z\right)}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz+1+z}{1+xz+z}=1\)
Vậy D = 1
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
Ta có : A = \(\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{xz+z+1}\)
A = \(\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\) (Vì xyz = 2017)
A = \(\dfrac{xy\left(xz\right)}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)
A = \(\dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}\)
A = \(\dfrac{xz+1+z}{xz+1+z}\) = 1
Vậy A = 1
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
\(\dfrac{xy}{x+y}=\dfrac{yz}{y+z}=\dfrac{zx}{z+x}\\ \Rightarrow\dfrac{x+y}{xy}=\dfrac{y+z}{yz}=\dfrac{z+x}{zx}\\ \Rightarrow\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{1}{z}+\dfrac{1}{y}=\dfrac{1}{x}+\dfrac{1}{z}\\ \Rightarrow\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}\\ \Rightarrow x=y=z\)
\(\Rightarrow P=\dfrac{xy+yz+zx}{x^2+y^2+z^2}=\dfrac{x^2+x^2+x^2}{x^2+x^2+x^2}=1\)
thay xyz=2017, ta có:
\(D=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xzy}+\frac{z}{xz+z+1}\)
\(D=\frac{xz}{1+xz+z}+\frac{1}{x+1+xz}+\frac{z}{xz+x+1}=1\)
\(\text{Bài làm }\)
\(\text{ Gọi xyz = 2017}\)
\(\text{Ta có:}\) \(D=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xzy}+\frac{z}{xz+z+1}\)
\(D=\frac{xz}{1+xz+z}+\frac{1}{x+1+xz}+\frac{z}{xz+x+1}=1\)
\(\text{# Chúc bạn học tốt #}\)