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Tìm các số hữu tỉ dương x,y,z biết:
\(\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=1-\frac{1}{2+\frac{1}{3}}\)
(x^2-2+1/x^2 ) +( y^2-2+1/y^2) +(z^2-2+1/z^2) =0
=> (x-1/x)^2 +(y-1/y)^2+(z-1/z)^2=0
suy ra x-1/x=0
y-1/y=0
z-1/z=0
.....
Ta có: \(x^2+\frac{1}{x^2}\ge2\sqrt{x^2.\frac{1}{x^2}}=2\)
\(y^2+\frac{1}{y^2}\ge2\sqrt{y^2.\frac{1}{y^2}}=2\)
\(z^2+\frac{1}{z^2}\ge2\sqrt{x^2.\frac{1}{z^2}}=2\)
\(\Rightarrow VT\ge6\)
Dấu "=" khi \(\orbr{\begin{cases}x=y=z=1\\x=y=z=-1\end{cases}}\)
\(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{\left(y+z+x+z+x+y\right)+\left(1+2-3\right)}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{2x+2y+2x}{x+y+z}\)=\(\frac{1}{x+y+z}\)
2=\(\frac{1}{x+y+z}\)(1)
Từ(1) => \(\frac{1}{x+y+z}\)=2 => x+y+z=0,5=>x+z=0,5-y(2)
Từ(1)=> x+y+1=2x(3)
x+z+2=2y(4)
z+y-3=2z(5)
Thay(2) vào (4) ta được: 0,5-y+2=2y
=> 2,5=3y
=> y=\(\frac{5}{6}\)
Thay y=\(\frac{5}{6}\)vào(3) ta được:x+\(\frac{5}{6}\)+1=2x
\(\frac{11}{6}\)=x
Thay x=\(\frac{11}{6}\); y=\(\frac{5}{6}\)vào x+y+z=0,5 ta đươc:
\(\frac{11}{6}\)+\(\frac{5}{6}\)+z=0,5
z=\(\frac{-13}{6}\)
Vậy ............
chúc bn học tốt.
k cho mik nha
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\)\(\frac{x+y-3}{z}=\frac{2\left(x+y+z\right)}{x+y+z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\frac{y+z+1}{x}+1=\frac{\frac{3}{2}}{x}=3\Rightarrow x=\frac{1}{2}\)
Tương tự suy ra \(y=\frac{5}{6},z=-\frac{5}{6}\)
k cho mình nha!
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow y+z=\frac{1}{2}-x;x+z=\frac{1}{2}-y;z+y=\frac{1}{2}-x\)
THAY VÀO BIỂU THỨC TA CÓ:
\(\frac{\frac{1}{2}-x+1}{x}=2\Rightarrow\frac{3}{2}-x=2x\Rightarrow x=\frac{1}{2}\)
\(\frac{\frac{1}{2}-y+2}{y}=2\Rightarrow\frac{5}{2}-y=2y\Rightarrow y=\frac{5}{6}\)
\(\frac{\frac{1}{2}-z-3}{z}=2\Rightarrow\frac{-5}{2}-z=2z\Rightarrow z=-\frac{5}{6}\)
\(\frac{y+z+1}{x}+\frac{x+z+2}{y}+\frac{x+y-3}{z}=\frac{y+x+1+x+z+2+x+y-3}{x+y+x}=\frac{2x+2y+2z}{x+y+z}=2.\)
\(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}=0,5\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\)\(\Rightarrow\frac{y+z+1}{x}+1=\frac{x+z+2}{y}+1=\frac{x+y-3}{z}+1=0,5+1\)
\(\Leftrightarrow\frac{x+y+z+1}{x}=\frac{x+y+z+2}{y}=\frac{x+y+z-3}{z}=1,5\)
\(\Leftrightarrow\frac{0,5+1}{x}=\frac{0,5+2}{y}=\frac{0,5-3}{z}=1,5\)
\(\Rightarrow\hept{\begin{cases}\frac{1,5}{x}=1,5\\\frac{2,5}{y}=1,5\\\frac{-2,5}{z}=1,5\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=1,6\\z=-1,6\end{cases}}}\)