help me

câu 1 là cm hay là tìm n

Ta có:(x+y):(5-z):(y+z):(y+9)=3:1:2:5

=> 5-z=1=>z=4.

    y+9=5=>y=-4.

   x+y=3=>x-4=3(do y=-4)=>x=7.

Vậy x=7,y=-4,z=4.

\(x+y+z=\frac12\left(x+y+y+z+x+z\right)\)

\(=\frac12\left(\frac12+\frac13+\frac14\right)=\frac12\cdot\frac{6+4+3}{12}=\frac12\cdot\frac{13}{12}=\frac{13}{24}\)

\(x+y+z=\frac{13}{24}\)

=>\(z+\frac12=\frac{13}{24}\)

=>\(z=\frac{13}{24}-\frac12=\frac{13}{24}-\frac{12}{24}=\frac{1}{24}\)

\(x+y+z=\frac{13}{24}\)

=>\(y+\frac14=\frac{13}{24}\)

=>\(y=\frac{13}{24}-\frac14=\frac{13}{24}-\frac{6}{24}=\frac{7}{24}\)

\(x+y+z=\frac{13}{24}\)

=>\(x+\frac13=\frac{13}{24}\)

=>\(x=\frac{13}{24}-\frac13=\frac{13}{24}-\frac{8}{24}=\frac{5}{24}\)

                  Giải

Ta có : \(\hept{\begin{cases}x\left(x+y+z\right)=-5\\y\left(x+y+z\right)=9\\z\left(x+y+z\right)=5\end{cases}}\Rightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=-5+9+5\)

\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=9\)

\(\Rightarrow\left(x+y+z\right)^2=3^2\)

\(\Rightarrow x+y+z=3\)

\(\Rightarrow\hept{\begin{cases}3x=-5\\3y=9\\3z=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-5}{3}\\y=3\\z=\frac{5}{3}\end{cases}}\)

Mà x , y , z là các số nguyên nên không có nghiệm x , y , z cần tìm