Áp dụng t/c dtsbn ta có:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\dfrac{1}{x+y+z}=2\Rightarrow2x+2y+2z=1\Rightarrow x+y+z=0,5\Rightarrow\left\{{}\begin{matrix}x+y=0,5-z\\y+z=0,5-x\\x+z=0,5-y\end{matrix}\right.\\ \dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\Rightarrow0,5-x+1=2x\Rightarrow x=0,5\\ \dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\Rightarrow0,5-y+2=2y\Rightarrow y=\dfrac{5}{6}\\ \dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\Rightarrow0,5-z-3=2z\Rightarrow z=-\dfrac{5}{6}\)

đúng rùi đó

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

\(\dfrac{x}{2}=\dfrac{y}{3}\text{⇒}\dfrac{x}{10}=\dfrac{y}{15}\)

\(\dfrac{y}{5}=\dfrac{z}{4}\text{⇒}\dfrac{y}{15}=\dfrac{z}{12}\)

⇒\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

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

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-21}{-3}=7\)

⇒x=70;y=105;z=84

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)⇒\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}\)

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

\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}=\dfrac{x^2-2y^2+z^2}{4-18+25}=\dfrac{44}{11}=4\)

⇒x=8;y=12;z=20

a) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/3 = y/4 = x/3 + y/4 = 28/7 = 4

=> x = 4 × 3 = 12

=> y = 4 × 4 = 16

Vậy x = 12, y = 16

B) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/2 = y/(-5) = x/2 - y/(-5) = (-7)/7 = -1

=> x = -1 × 2 = -2

=> y = -1 × -5 = 5

Vậy x = -2, y = 5

C) làm tương tự như bài a, b

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

Do đó: x=16; y=24; z=30

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

Do đó: x=16; y=24; z=30

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z=\dfrac{x+y+z}{y+z+1+x+z+1+x+y-2}=\dfrac{x+y+z}{2x+2y+2z}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)

\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\Rightarrow y+z+1=2x\Rightarrow y+z=2x-1\left(1\right)\)

\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\Rightarrow x+z+1=2y\Rightarrow x+z=2y-1\left(2\right)\)

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

\(x+y+z=\dfrac{1}{2}\left(3\right)\)

Thay (1) vào (3) ta có:

\(x+y+z=\dfrac{1}{2}\\ \Rightarrow x+2x-1=\dfrac{1}{2}\\ \Rightarrow3x=\dfrac{3}{2}\\ \Rightarrow x=\dfrac{1}{2}\)

Thay (2) vào (3) ta có:

\(x+y+z=\dfrac{1}{2}\\ \Rightarrow y+2y-1=\dfrac{1}{2}\\ \Rightarrow3y=\dfrac{3}{2}\\ \Rightarrow y=\dfrac{1}{2}\)

Ta có:

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

TH1: \(x+y+z=0\Rightarrow x=y=z=0\)

TH2: \(x+y+z\ne0\)

\(x+y+z=\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x+2y+2z=1\\2x=y+z+1\\2y=x+z+1\\2z=x+y-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x+2y+2z=1\\2x+2y+2z=3y+3z+1\\2x+2y+2z=3x+3z+1\\2x+2y+2z=3x+3y-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y+2z=1\\y+z=0\\x+z=0\\x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2.1+2z=1\\y=-z\\x=-z\\x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}z=-\dfrac{1}{2}\\x=\dfrac{1}{2}\\y=\dfrac{1}{2}\\\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(0;0;0\right);\left(\dfrac{1}{2};\dfrac{1}{2};-\dfrac{1}{2}\right)\)

Lời giải:
Áp dụng TCDTSBN:

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

\(\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ y+z+1=2x\\ x+z+2=2y\\ x+y-3=2z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ x+y+z+1=3x\\ x+y+z+2=3y\\ x+y+z-3=3z\end{matrix}\right.\)

\(\left\{\begin{matrix} \frac{1}{2}+1=3x\\ \frac{1}{2}+2=3y\\ \frac{1}{2}-3=3z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=\frac{1}{2}\\ y=\frac{5}{6}\\ z=\frac{-5}{6}\end{matrix}\right.\)