1. áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{x+2}{3}=\frac{y-7}{5}=\frac{x+y-5}{3+5}=\frac{16}{8}=2\Rightarrow\hept{\begin{cases}x+2=6\\y-7=10\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=17\end{cases}}}\)

2. áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-y+2}{2-3}=\frac{-10+7}{-1}=3\Rightarrow\hept{\begin{cases}x+5=6\\y-2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=11\end{cases}}\)

Ta có : \(\frac{x}{5}=y=\frac{z}{-2}\Rightarrow\frac{x}{5}=\frac{y}{1}=\frac{z}{-2}\Rightarrow\frac{x}{5}=\frac{y}{1}=\frac{2z}{-4}\)

Lại có : -x - y + 2z = 160

=> -(x + y - 2z) = 160

=> x + y - 2z = -160

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

\(\frac{x}{5}=\frac{y}{1}=\frac{2z}{-4}=\frac{x+y-2z}{5+1-\left(-4\right)}=\frac{-160}{10}=-16\)

=> x = -16.5 = -80 , y = -16 , z = -16.(-2) = 32

Đặt \(\frac{x}{3}=\frac{y}{8}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=3k\\y=8k\\z=5k\end{cases}}\)

=>  4x = 12k , 3y = 24k , 2z = 10k

=> 4x + 3y - 2z = 12k + 24k - 10k

=> 52 = 26k

=> k = 2

Với k = 2 thì x = 3.2 = 6 , y = 8.2 = 16 , z=  5.2 = 10

8x = 5y => \(\frac{x}{5}=\frac{y}{8}\)

=> \(\frac{2x}{10}=\frac{y}{8}\)

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

\(\frac{2x}{10}=\frac{y}{8}=\frac{y-2x}{8-10}=\frac{-10}{-2}=5\)

=> x = 5.5 = 25,y = 5.8 = 40

\(\dfrac{x}{2}=\dfrac{z}{3};\dfrac{y}{5}=\dfrac{z}{2}\Rightarrow\dfrac{x}{4}=\dfrac{z}{6}=\dfrac{y}{15}\)

Theo tc dãy tỉ số bằng nhau 

\(\dfrac{x}{4}=\dfrac{z}{6}=\dfrac{y}{15}=\dfrac{x+y+z}{4+6+15}=\dfrac{50}{25}=2\Rightarrow x=8;y=12;y=30\)

Ta có : \(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)

\(\Rightarrow\left(\frac{x}{5}\right)^2=\left(\frac{y}{7}\right)^2=\left(\frac{z}{3}\right)^2=\frac{x^2}{5^2}=\frac{y^2}{7^2}=\frac{z^2}{3^2}\)\(=\frac{x^2}{25}=\frac{y^2}{49}=\frac{z^2}{9}=\frac{x^2+y^2-z^2}{25+49-9}=\frac{585}{65}=9\)

\(\Rightarrow x=9.5=45\)

     \(y=9.7=63\)

     \(z=9.3=27\)

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)