\(x+y+z=\frac{x}{y+z-2}=\frac{y}{z+x-3}=\frac{z}{x+y+5}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{y+z-2}=\frac{y}{z+x-3}=\frac{z}{x+y+5}=\frac{x+y+z}{2\left(x+y+z\right)-2-3+5}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\Rightarrow z+y+z=\frac{1}{2}\)Ta có:

\(\frac{x}{y+z+1}=\frac{1}{2}\)

\(\Rightarrow2x=y+z+1\)

\(\Rightarrow y+z=2x-1\)

\(\Rightarrow x+\left(2x-1\right)=\frac{1}{2}\)

\(\Rightarrow x+2x-1=\frac{1}{2}\)

\(\Rightarrow3x-1=\frac{1}{2}\)

\(\Rightarrow3x=\frac{1}{2}+1\)

\(\Rightarrow3x=\frac{3}{2}\)

\(\Rightarrow x=\frac{3}{2}:3\)

\(\Rightarrow x=\frac{1}{2}\)

y ;z bạn làm tương tự

- Mình nhầm chỗ \(\frac{x}{y+z+1}\)tí sữa thành \(\frac{x}{y+z+2}\)nhá D

1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)

\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)

=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)

=>\(x=3\cdot20=60\)

    \(y=3\cdot24=72\)

    \(z=3\cdot21=63\)

3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)

=> \(x=1\cdot15=15\)

     \(y=1\cdot7=7\)

     \(z=1\cdot3=3\)

     \(t=1\cdot1=1\)

??????,,,,,

ai mua , đổi acc bang bang thì nhắn tin vs tui

\(\frac{y+z+2}{x}=\frac{x+z+3}{y}=\frac{x+y-5}{z}=\frac{1}{x+y+z}\)

=>\(\frac{\left(x+y+z\right)2}{x+y+z}=\frac{1}{x+y+z}\)

=> x+y+z=1/2

=> y+z=2x-2

=>    x+z=2y-3

=>x+y=2x+5

=> 1/2-x=2x-3

=> x=5/6

=>1/2-y=2y-3

=> y=7/6

=> z=1/2-(7/6+5/6)=-3/2

chtt