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Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+y+2015}{z}=\frac{y+z-2016}{x}=\frac{z+x+1}{y}.\)
\(=\frac{x+y+2015+y+z-2016+z+x+1}{x+y+z}\)\(=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Do đó x+y+z=1 => x+y=1-z => \(\frac{2016-z}{z}=2\Rightarrow2016-z=2z\Leftrightarrow2016=3z\)
=> z= 672
Tương tự : x= -2015/3; y=2/3
Ta có:
\(\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(=1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}=3+\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)
\(\Rightarrow x+y+z=\frac{3+\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)}{\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}}=\frac{3+\frac{7}{10}}{\frac{2}{5}}=\frac{37}{4}\)
Ta có :
\(\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+x}+\frac{1}{z+x}\right)\)
\(=1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}=3+\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)
\(\Rightarrow x+y+z=\frac{3+\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)}{\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}}=\frac{3+\frac{7}{10}}{\frac{2}{5}}=\frac{37}{4}\)
Ta có :
\(x+y=\frac{1}{2}\)
\(y+z=\frac{1}{3}\)
\(z+x=\frac{1}{4}\)
\(\Rightarrow\)\(x+y+y+z+z+x=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\Rightarrow\)\(2x+2y+2z=\frac{13}{12}\)
\(\Rightarrow\)\(2\left(x+y+z\right)=\frac{13}{12}\)
\(\Rightarrow\)\(x+y+z=\frac{13}{12}:2\)
\(\Rightarrow\)\(x+y+z=\frac{13}{24}\)
Do đó :
\(x+y+z=\frac{13}{24}\)
\(\Rightarrow\)\(x=\frac{13}{24}-\left(y+z\right)=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)
\(\Rightarrow\)\(y=\frac{13}{24}-\left(z+x\right)=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\)
\(\Rightarrow\)\(z=\frac{13}{24}-\left(x+y\right)=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)
Vậy \(x=\frac{5}{24};y=\frac{7}{24};z=\frac{1}{24}\)
Chúc bạn học tốt ~