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Áp dụng t/c dtsbn:
\(\dfrac{x}{33}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{z-x}{5-33}=\dfrac{-196}{-28}=7\)
\(\Rightarrow\left\{{}\begin{matrix}x=7.33=231\\y=7.4=28\\z=7.5=35\end{matrix}\right.\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}=\dfrac{-x+z}{-\dfrac{11}{6}+\dfrac{5}{18}}=\dfrac{-196}{-\dfrac{14}{9}}=126\)
Do đó: x=231; y=28; z=35
\(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}z\)
\(\Leftrightarrow\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}=\dfrac{x+y+z}{\dfrac{11}{6}+\dfrac{2}{9}+\dfrac{5}{18}}=\dfrac{-196}{\dfrac{42}{18}}=\dfrac{-98}{\dfrac{21}{18}}=\dfrac{-588}{7}\)
(thấy lẻ,nếu đề ko sai thì làm tiếp)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
\(=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x=2y\\2z=4x\\4y=3z\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=\dfrac{-10}{1}=-10\)
\(\Rightarrow\left\{{}\begin{matrix}x=-10.2=-20\\y=-10.3=-30\\z=-10.4=-40\end{matrix}\right.\)
Vậy......
a.
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)
Thế vào \(2x+y-z=81\)
\(\Rightarrow2.5k+3k-4k=81\)
\(\Rightarrow9k=81\)
\(\Rightarrow k=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)
b.
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)
Thế vào \(5x-y+3z=124\)
\(\Rightarrow5.3k-5k+3.2k=124\)
\(\Rightarrow16k=124\)
\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)
c.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Thế vào \(xyz=810\)
\(\Rightarrow2k.3k.5k=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)
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
1) Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).3=-3\\y=\left(-1\right).5=-5\\z=\left(-1\right).7=-7\end{matrix}\right.\)
2) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{19}.8=-\dfrac{224}{19}\\y=-\dfrac{28}{19}.12=-\dfrac{336}{19}\\z=-\dfrac{28}{19}.15=-\dfrac{420}{19}\end{matrix}\right.\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{3\cdot2+5\cdot3-7}=\dfrac{-14}{14}=-1\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
b, \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{224}{19}\\y=-\dfrac{336}{19}\\z=-\dfrac{420}{19}\end{matrix}\right.\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{33}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{z-x}{5-33}=\dfrac{-196}{-28}=7\)
Do đó: 231; y=28; z=35