Tim ba so x, y, z biet \(\frac{y+z+1}{x}=\frac{x+ z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
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Tim x , y , z biet:
\(\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{z}{x+y+3}=x+y+z\)
Tim x,y,z biet rang: \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
tim x,y,z biet: \(\frac{x+z+2}{y}=\frac{y+z+1}{x}=x+y+3=\frac{1}{x+y+z}\)
tim x,y,z biet
\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z\)
Đặt \(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z=k\)
Áp dụng TC DTSBN ta có :
\(k=\frac{x+y+z}{\left(y+z+1\right)+\left(z+x+1\right)+\left(x+y-2\right)}=\frac{\left(x+y+z\right)}{2\left(x+y+z\right)}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}y+z+1=2x\\z+x+1=2y\\x+y-2=2z\end{cases}}\) và \(x+y+z=\frac{1}{2}\)
\(\Leftrightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+1=3y\\x+y+z-2=3z\end{cases}}\) và \(x+y+z=\frac{1}{2}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{2}+1=3x\\\frac{1}{2}+1=3y\\\frac{1}{2}-2=3z\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\\-\frac{1}{2}\end{cases}}\)
Vậy \(x=\frac{1}{2};y=\frac{1}{2};z=-\frac{1}{2}\)
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Tim x , y , z biet:
\(\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{z}{x+y-3}=x+y+z\)
Cach lam ho minh voi
Tim x;y;z biet:
\(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)
ta có:\(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{y+x-2}=\frac{x+y+z}{2\left(x+y+x\right)}=\frac{1}{2}\)
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Tim x,y,z biet:
a, xy=z ; yz=4x ; zx=9x
b, \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(\frac{y+z+1+x+z+1+x+y-3}{x+y+z}\)=\(\frac{2\left(X+Y+Z\right)}{x+y+z}\)=2 =>x+y+z=\(\frac{1}{2}\) tu lam di nhe
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Tim x,y,z biet:
a, \(xy=z;yz=4x;zx=9y\)
b, \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
Tim cap so x;y;z biet :
\(\frac{y+x+1}{x}\)=\(\frac{x+z+2}{y}\)=\(\frac{x+y-3}{z}\)=\(\frac{1}{x+y+z}\)