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a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\)(do 1/(x+y+z)=2)
\(\Rightarrow y+z=\frac{1}{2}-x;z+x=\frac{1}{2}-y;x+y=\frac{1}{2}-z\)
Thay vào lần lượt ta có:
\(\frac{\frac{1}{2}-x+1}{x}=2\)\(\Rightarrow x=\frac{1}{2}\)
\(\frac{\frac{1}{2}-y+2}{y}=2\)\(\Rightarrow y=\frac{5}{6}\)
\(\frac{\frac{1}{2}-z-3}{z}=2\)\(\Rightarrow z=-\frac{5}{6}\)
b) \(\frac{x-1}{2}=\frac{2x-2}{4}\)
\(\frac{y-2}{3}=\frac{3y-6}{9}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z+3-2-6}{9}=\frac{50+3-2-6}{9}=\frac{45}{9}=5\)=>x-1=5.2=10
=>x=11
y-2=5.3=15
=>y=17
z-3=5.4=20
=>z=23
Vậy (x;y;z)=(11;17;23)
Áp dụng t/c của dãy tỉ số bằng nhau:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+x-3\right)}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)(vì x+y+z khác 0).Do đó x+y+z = 0.5
Thay kq này vào bài ta được:
\(\frac{0,5-x+1}{x}=\frac{0,5-y+2}{y}=\frac{0,5-z-3}{z}=2\)
Tức là : \(\frac{1,5-x}{x}=\frac{2,5-y}{y}=\frac{-2,5-z}{z}=2\)
Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=\frac{-5}{6}\)
\(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{\left(y+z+x+z+x+y\right)+\left(1+2-3\right)}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{2x+2y+2x}{x+y+z}\)=\(\frac{1}{x+y+z}\)
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
Ta có: \(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=x+y+z\)
TH1: \(x+y+z=0\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{x+y+z}{x+y-3+y+z+1+z+x+2}\)
\(=\frac{x+y+z}{x+y+y+z+z+x}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
\(\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow x+y=\frac{1}{2}-z\)
\(y+z=\frac{1}{2}-x\)
\(z+x=\frac{1}{2}-y\)
Thay \(x+y-3=\frac{1}{2}-z-3\)
\(\Rightarrow\frac{z}{\frac{1}{2}-z+3}=\frac{1}{2}\)
\(\Rightarrow2z=\frac{1}{2}-z-3\)
\(\Rightarrow2z+z=\frac{1}{2}-3\)
\(\Rightarrow3z=-\frac{5}{2}\Rightarrow z=-\frac{5}{6}\)
Thay \(y+z+1=\frac{1}{2}-x+1\)
\(\Rightarrow\frac{x}{\frac{1}{2}-x+1}=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}-x+1\)
\(\Rightarrow2x+x=\frac{1}{2}+1\)
\(\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(z+x+2=\frac{1}{2}-y+2\)
\(\Rightarrow\frac{y}{\frac{1}{2}-y+2}=\frac{1}{2}\)
\(\Rightarrow2y=\frac{1}{2}-y+2\)
\(\Rightarrow2y+y=\frac{1}{2}+2\)
\(\Rightarrow3y=\frac{5}{2}\Rightarrow y=\frac{5}{6}\)
Ta có: \(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)
\(=\left(\frac{1}{2}+\frac{5}{6}+-\frac{5}{6}-\frac{3}{2}\right)^{2019}\)
\(=\left[\left(\frac{1}{2}-\frac{3}{2}\right)+\left(-\frac{5}{6}+\frac{5}{6}\right)\right]^{2019}\)
\(=\left(-1\right)^{2019}=-1\)
TH2: x + y + z = 0
\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=0\)
\(\Rightarrow x=y=z=0\)
\(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)
\(=\left(0-\frac{3}{2}\right)^{2019}=\left(-\frac{3}{2}\right)^{2019}\)
Ah! Mk nhầm chút. TH1 là khác 0 nhé!!!!!!
Điều kiện \(\hept{\begin{cases}x\ne0\\y\ne0\\z\ne0\end{cases}}\)
ADTC dãy tỉ số bằng nhau ta có :
\(\frac{\left(y+z+1\right)}{x}=\frac{\left(x+z+2\right)}{y}=\frac{\left(x+y-3\right)}{z}=\downarrow\)
\(=\frac{\left(y+z+1+x+z+2+x+y-3\right)}{\left(x+y+z\right)}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\frac{1}{\left(x+y+z\right)}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\Leftrightarrow y+z=\frac{1}{2}-x\)(1)
\(\frac{\left(y+z+1\right)}{x}=2\Leftrightarrow y+z+1=2x\)
Kết hợp với (1) \(\Rightarrow\frac{1}{2}-x+1=2x\)
\(\Leftrightarrow x=\frac{1}{2}\Rightarrow y+z=0\Leftrightarrow y=-z\)
Ta có : \(\frac{\left(x+y-3\right)}{z}=2\)
\(\Leftrightarrow x+y-3=2z\)
\(\Leftrightarrow y-2z=\frac{5}{2}\)
Do: \(y=-z\Rightarrow-3z=\frac{5}{2}\Leftrightarrow z=-\frac{5}{6}\)
\(\Rightarrow y=\frac{5}{6}\)
Vậy nghiệm tìm đc : \(\left(x;y;z\right)=\left(\frac{1}{2};\frac{5}{6};-\frac{5}{6}\right)\)