tim x,y,z biet
\(\frac{x+y+2019}{z}\)=\(\frac{y+z-2020}{x}\)=\(\frac{z+x+1}{y}\)=\(\frac{2}{x+y+z}\)
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 \(x=\frac{1}{2};y=\frac{1}{2};z=-\frac{1}{2}\)
Tìm x;y;z thỏa mãn:
\(\frac{\sqrt{x-2018}-1}{x-2018}+\frac{\sqrt{y-2019}-1}{y-2019}+\frac{\sqrt{z-2020}-1}{z-2020}=\frac{3}{4}\)
Tim x , y , z biet:
\(\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{z}{x+y+3}=x+y+z\)
Tim ba so x, y, z biet \(\frac{y+z+1}{x}=\frac{x+ z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
Suy ra
\(x+y+z=\frac{1}{2}\)(1)
\(y+z+1=2x\)(2)
\(x+z+2=2y\)(3)
\(x+y-3=2z\)(4)
(2)-(1) ta có
\(1-x=2x-\frac{1}{2}\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)
\(x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\Leftrightarrow y+z=\frac{1}{2}-\frac{1}{2}=0\)
\(y=-z\)
\(x+z+2=\frac{1}{2}+2-y==\frac{5}{2}-y\)
\(\frac{\frac{5}{2}-y}{y}=\frac{5}{2y}-1=2\Leftrightarrow\frac{5}{2y}=3\Leftrightarrow y=\frac{5}{6}\)
\(z=-\frac{5}{6}\)
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:
\(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)
ta có:\(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{y+x-2}=\frac{x+y+z}{2\left(x+y+x\right)}=\frac{1}{2}\)
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
Cho các số a,b,c,d khác 0 và x,y,z,t thỏa mãn :
\(\frac{x^{2020}+y^{2020}+z^{2020}+t^{2020}}{a^{2020}+b^{2020}+c^{2020}+d^{2020}}=\frac{x^{2020}}{a^{2020}}+\frac{y^{2020}}{b^{2020}}+\frac{z^{2020}}{c^{2020}}+\frac{t^{2020}}{d^{2020}}\)
Tính \(T=x^{2019}+y^{2019}+z^{2019}+t^{2019}\)
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