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a) Với \(x+y+z=0\) ta tìm được \(\left(x;y;z\right)\rightarrow\left(0;0;0\right)\)
Với \(x+y+z\ne0\) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{z+x+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)
Hay: \(x+y+z=\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}y+z=\dfrac{1}{2}-x\\x+z=\dfrac{1}{2}-y\\x+y=\dfrac{1}{2}-z\end{matrix}\right.\)
Thay vào đề bài ta được:
\(\dfrac{x}{\dfrac{1}{2}-x+1}=\dfrac{y}{\dfrac{1}{2}-y+1}=\dfrac{z}{\dfrac{1}{2}-z-2}=\dfrac{1}{2}\) Dễ dàng tìm được x;y;z
b) Theo đề bài ta có sẵn x+y+z khác 0
Áp dụng dãy tỉ số rồi làm tương tự câu a
Áp dụng t/c dãy t/s = nhau:
\(\frac{y+x+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\frac{2.\left(x+y+z\right)}{x+y+z}=2\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=2\)
\(\Rightarrow y+z+1=2x\)
\(x+z+2=2y\)
\(x+y-3=2z\)
\(x+y+z=\frac{1}{2}\)
*) \(x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\)Thay vào \(y+z+1=2x\)ta được \(\frac{1}{2}-x+1=2x\Rightarrow\frac{3}{2}=3x\Rightarrow x=\frac{1}{2}\)
*) \(x+y+z=\frac{1}{2}\Rightarrow x+z=\frac{1}{2}-y\) Thay vào \(x+z+2=2y\) ta được \(\frac{1}{2}-y+2=2y\Rightarrow\frac{5}{2}=3y\Rightarrow y=\frac{5}{6}\)
\(\Rightarrow x+y+z=\frac{1}{2}+\frac{5}{6}+z=\frac{1}{2}\Rightarrow\frac{4}{3}+z=\frac{1}{2}\Rightarrow z=\frac{1}{2}-\frac{4}{3}=-\frac{5}{6}\)
a/ Ta có ;
\(x+y+z=92\)
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{7}\)
\(\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{10}=2\Leftrightarrow x=20\\\dfrac{y}{15}=2\Leftrightarrow y=30\\\dfrac{z}{21}=2\Leftrightarrow z=42\end{matrix}\right.\)
Vậy .................
b/Ta có :
\(x+y-z=95\)
\(2x=3y=5z\)
\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\)
\(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{5}\)
Áp dụng t/x dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{5}=\dfrac{x+y-z}{15+10-5}=\dfrac{95}{19}=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=5\Leftrightarrow x=75\\\dfrac{y}{10}=5\Leftrightarrow y=50\\\dfrac{z}{5}=5\Leftrightarrow z=25\end{matrix}\right.\)
Vậy ..
a, \(\dfrac{x}{2}=\dfrac{y}{3},\dfrac{y}{5}=\dfrac{z}{7},x+y+z=92\)
Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{15}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{21}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21},x+y+z=92\)
AD t/c DTS = nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)
+) \(\dfrac{x}{10}=2\Rightarrow x=20\)
+) \(\dfrac{y}{15}=2\Rightarrow y=30\)
+) \(\dfrac{z}{21}=2\Rightarrow z=42\)
b, \(2x=3y=5z,x+y-z=95\)
\(\Rightarrow\dfrac{30x}{15}=\dfrac{30y}{10}=\dfrac{30z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6},x+y-z=95\)
AD t/c DTS = nhau ta có:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
+) \(\dfrac{x}{15}=5\Rightarrow x=75\)
+) \(\dfrac{y}{10}=5\Rightarrow y=50\)
+) \(\dfrac{z}{6}=5\Rightarrow z=30\)
c, Bn xem lại đề bài nha! 
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2.\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\\\dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\\\dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\\\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow y+z=\dfrac{1}{2}-x\). Thay vào \(y+z+1=2x\) ta được \(\dfrac{1}{2}-x+1=2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+z=\dfrac{1}{2}-y\). Thay vào \(x+z+2=2y\) ta được \(\dfrac{1}{2}-y+2=2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)
\(\Rightarrow x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+z=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}+z=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}-\dfrac{4}{3}=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2};y=\dfrac{5}{6};z=\dfrac{-5}{6}\)
Lời giải:
ĐK: \(x,y,z,x+y+z\neq 0\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+z+x+2+x+y-3}{x+y+z}=\frac{2(x+y+z)}{x+y+z}\)
\(\Rightarrow \frac{1}{x+y+z}=\frac{2(x+y+z)}{x+y+z}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\)
Do đó thay vào điều kiện đề bài ban đầu:
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow \frac{\frac{1}{2}-x+1}{x}=\frac{\frac{1}{2}-y+2}{y}=\frac{\frac{1}{2}-z-3}{z}=2\)
\(\Leftrightarrow \frac{3}{2x}-1=\frac{5}{2y}-1=\frac{-5}{2z}-1=2\)
\(\Leftrightarrow \frac{3}{2x}=\frac{5}{2y}=\frac{-5}{2z}=3\)
\(\Rightarrow x=\frac{1}{2}; y=\frac{5}{6}; z=\frac{-5}{6}\)
Thử lại thấy đúng.
Vậy.................
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Theo bài ra: \(\dfrac{x+y-3}{z}=\dfrac{x+z+2}{y}=\dfrac{y+z+1}{x}=\dfrac{1}{x+y+z}\Rightarrow\dfrac{x+y-3}{z}=\dfrac{x+z+2}{y}=\dfrac{y+z+1}{x}=\dfrac{1}{x+y+z}=2\)
\(\Rightarrow x+y-3=2z\left(1\right);x+z+2=2y\left(2\right);y+z+1=2x\left(3\right);x+y+z=\dfrac{1}{2}\)
\(+)x+y+z=\dfrac{1}{2}\Leftrightarrow y+x=\dfrac{1}{2}-z\). Thay vào \(\left(3\right)\), ta được\(\dfrac{1}{2}-x+1=2x\Rightarrow\dfrac{3}{2}=3x\Rightarrow x=\dfrac{1}{2}\)
\(+)x+y+z=\dfrac{1}{2}\Leftrightarrow x+z=\dfrac{1}{2}-y\). Thay vào \(\left(2\right)\), ta được \(\dfrac{1}{2}-y+2=2y\Rightarrow\dfrac{5}{2}=3y\Rightarrow y=\dfrac{5}{6}\)
\(\Rightarrow x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+z\Leftrightarrow z=-\dfrac{5}{6}\)
Vậy \(x=\dfrac{1}{2};y=\dfrac{5}{6};z=-\dfrac{5}{6}\)
Đặt \(\dfrac{y+x+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=k\)
Áp dụng TC DTSBN ta có :
\(k=\dfrac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\dfrac{2x+2y+2z}{x+y+z}=2\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{2};y=\dfrac{5}{6};z=-\dfrac{5}{6}\)
Hình như sai đề rồi bạn :
Có phải như thế này không :
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+y}\)
Ta có\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}\)
\(=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{2x+2y+2z+1+2-3}{x+y+z}\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Nên \(\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)
Ta lại có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=2\)
\(\Leftrightarrow\dfrac{\left(x+y+z\right)-z+1}{x}=\dfrac{\left(x+y+z\right)-y+2}{y}=\dfrac{\left(x+y+z\right)-z-3}{z}=2\)
\(\Rightarrow\dfrac{\dfrac{1}{2}-x+1}{x}=\dfrac{\dfrac{1}{2}-y+2}{y}=\dfrac{\dfrac{1}{2}-z-3}{z}=2\)
\(\Rightarrow\dfrac{\dfrac{3}{2}-x}{x}=\dfrac{\dfrac{5}{2}-y}{y}=\dfrac{-z-\dfrac{5}{2}}{z}=2\)
\(\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\dfrac{3}{2}-x}{x}\\\dfrac{\dfrac{5}{2}-y}{y}\\\dfrac{-z-\dfrac{5}{2}}{z}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{2}-x\\2y=\dfrac{5}{2}-y\\2z=-z-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{5}{2}\end{matrix}\right.\)
Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau, với $x+y+z\neq 0$ ta có:
\(\frac{1}{x+y+z}=\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2(x+y+z)}{x+y+z}=2\)
\(\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ y+z+1=2x\\ x+z+2=2y\\ x+y-3=2z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ x+y+z+1=3x\\ x+y+z+2=3y\\ x+y+z-3=3z\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} \frac{1}{2}+1=3x\\ \frac{1}{2}+2=3y\\ \frac{1}{2}-3=3z\end{matrix}\right.\Rightarrow x=\frac{1}{2}; y=\frac{5}{6}; z=\frac{-5}{6}\)
Vậy..........