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

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{3\left(x-1\right)}{6}=\dfrac{3\left(y-2\right)}{9}=\dfrac{z-3}{4}\)

\(\Leftrightarrow\dfrac{3x-3}{6}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x-3}{6}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{3x-3+3y-6-z+3}{6+9-4}=\dfrac{\left(3x+3y-z\right)+\left(3-3-6\right)}{11}=\dfrac{50-6}{11}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=4\Leftrightarrow x=4.2+1=9\\\dfrac{y-2}{3}=4\Leftrightarrow y=4.3+2=14\\\dfrac{z-3}{4}=4\Leftrightarrow z=4.4+3=19\end{matrix}\right.\)

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) (1)

Áp dụng dãy tỉ số bằng nhau , ta có:

\(\left(1\right)\Leftrightarrow\dfrac{\left(x-1\right)+\left(y-2\right)-\left(z-3\right)}{2+3-4}\)

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

\(=x+y-z=50\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=50.2=100\\y-2=50.3=150\\z-3=50.4=200\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=101\\y=152\\z=203\end{matrix}\right.\)

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)

Do đó: x=-16; y=-24; z=-30

áp dụng tính chất dảy tỉ số bằng nhau

ta có : \(\dfrac{2\left(x-1\right)+3\left(y-2\right)-\left(z-3\right)}{\left(2.2\right)+\left(3.3\right)-4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}\)

\(=\dfrac{\left(2x+3y-z\right)-5}{9}=\dfrac{50-5}{9}=\dfrac{45}{9}=5\)

suy ra ta có : \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=2.5\\y-2=3.5\\z-3=4.5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=10+1\\y=15+2\\z=20+3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\) vậy \(x=11;y=17;z=23\)

\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)

\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(xyz=810\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

=> \(x=2k\) ; \(y=3k\) ; \(z=5k\)

Thay \(x=2k;y=3k;z=5k\) vào \(xyz=810\) ta được

\(2k.3k.5k=810\)

\(30k=810\)

\(k^3=27\)

=> k = 3

=> \(x=2.3=6\)

=> \(y=3.3=9\)

=> \(z=5.3=15\)

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

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

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

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

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

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

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

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

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

Thay vào \(y+z+1=2x\) ; ta có :

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

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

Thay vào \(x+z+2=2y\) ; ta có :

\(\dfrac{1}{2}-y+2=2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)

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

Thay vào \(x+y-3=2z\) ; ta có :

\(\dfrac{1}{2}-z-3=2z\Rightarrow3z=\dfrac{-5}{2}\Rightarrow z=\dfrac{-5}{6}\)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)

Áp dụng t/c dtsbn:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50+3-12-25}{8}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)

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