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a) Từ \(9x=3y=2z\) ta chia các vế cho 18 (là BCNN của 9, 3 và 2) ta được:
\(\frac{9x}{18}=\frac{3y}{18}=\frac{2z}{18}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{6}=\frac{z}{9}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{6}=\frac{z}{9}=\frac{x-y+z}{2-6+9}=\frac{50}{5}=10\)
=> \(\frac{x}{2}=10\Rightarrow x=10.2=20\)
\(\frac{y}{6}=10\Rightarrow y=10.6=60\)
\(\frac{z}{9}=10\Rightarrow z=10.9=90\)
b) Đặt \(k=\frac{x}{5}=\frac{y}{2}=\frac{z}{-3}\)
=> \(x=5k\) ; \(x=2k\) ; \(z=-3k\) (*)
Biết xyz = 240 => \(5k.2k.\left(-3k\right)=240\)
\(\Rightarrow-30k^3=240\)
\(\Rightarrow k^3=-8\)
\(\Rightarrow k=-2\)
Thay vào (*) ta được
\(x=5k=5.\left(-2\right)=-10\)
\(y=2k=-4\)
\(z=-3k=6\)
a)\(\hept{\begin{cases}9x=3y=2z\\x-y+z=50\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{9}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}\\x-y+z=50\end{cases}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{\frac{1}{9}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}=\frac{x-y+z}{\frac{1}{9}-\frac{1}{3}+\frac{1}{2}}=\frac{50}{\frac{5}{18}}=180\)
\(\Rightarrow\hept{\begin{cases}x=20\\y=60\\z=90\end{cases}}\)
b) Đặt \(\frac{x}{5}=\frac{y}{2}=\frac{z}{-3}=k\)
\(\Rightarrow\hept{\begin{cases}x=5k\\y=2k\\z=-3k\end{cases}}\)
xyz = 240 <=> 5k.2k.(-3)k = 240
<=> -30k3 = 240
<=> k3 = -8
<=> k3 = (-2)3
<=> k = -2
=> \(\hept{\begin{cases}x=-10\\y=-4\\z=6\end{cases}}\)
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1, \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)\(\Leftrightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=k\)\(\Leftrightarrow\hept{\begin{cases}x=2k\\y=\frac{3}{2}k\\z=\frac{4}{3}k\end{cases}}\)
Mà xyz = -108
\(\Leftrightarrow2k.\frac{3}{2}k.\frac{4}{3}k=-108\)
\(\Leftrightarrow4k^3=-108\)
<=> k3 = -27
<=> k = -3
\(\Leftrightarrow\hept{\begin{cases}x=2k=2.-3=-6\\y=\frac{3}{2}k=\frac{3}{2}.\left(-3\right)=\frac{-9}{2}\\z=\frac{4}{3}k=\frac{4}{3}.\left(-3\right)=-4\end{cases}}\)
2, \(\frac{x}{5}=\frac{y}{7}=\frac{z}{8}\)\(\Leftrightarrow\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}=\frac{2x+3y-4z}{10+21-32}=\frac{15}{-1}=-15\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=-15\\\frac{y}{7}=-15\\\frac{z}{8}=-15\end{cases}}\Rightarrow\hept{\begin{cases}x=-75\\y=-105\\z=-120\end{cases}}\)
3, 3x = 5y \(\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)\(\Leftrightarrow\frac{x}{55}=\frac{y}{33}\)
2y = 11z \(\Leftrightarrow\frac{y}{11}=\frac{z}{2}\) \(\Leftrightarrow\frac{y}{33}=\frac{z}{6}\)
\(\Rightarrow\frac{x}{55}=\frac{y}{33}=\frac{z}{6}\)\(\Rightarrow\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}=\frac{2x+5y-z}{110+165-6}=\frac{34}{269}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{55}=\frac{34}{269}\\\frac{y}{33}=\frac{34}{269}\\\frac{z}{6}=\frac{34}{269}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1870}{269}\\y=\frac{1122}{269}\\z=\frac{204}{269}\end{cases}}\)
4, \(\frac{x}{3}=\frac{2}{y}=\frac{z}{4}=k\)\(\Leftrightarrow\hept{\begin{cases}x=3k\\y=\frac{2}{k}\\z=4k\end{cases}}\)
Mà xyz = 240
<=> 3k . 2/k . 4k = 240
<=> 24k = 240
<=> k = 10
\(\Leftrightarrow\hept{\begin{cases}x=3k=3.10=30\\y=\frac{2}{k}=\frac{2}{10}=\frac{1}{5}\\z=4k=4.10=40\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{x}{4}=\frac{y}{3}=\frac{z}{5}=kak\left(kak\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=4kak\\y=3kak\\z=5kak\end{cases}}\)
Mà \(x^2+y^2+z^2=200\)
\(\Leftrightarrow\left(4kak\right)^2+\left(3kak\right)^2+\left(5kak\right)^2=200\)
\(\Leftrightarrow16.kak^2+9.kak^2+25.kak^2=200\)
\(\Leftrightarrow kak^2.\left(16+9+25\right)=200\)
\(\Leftrightarrow kak^2.50=200\)
\(\Leftrightarrow kak^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}kak=2\\kak=-2\end{cases}}\)
+) Với \(kak=2\)thì \(\hept{\begin{cases}x=4kak=8\\y=3kak=6\\z=5kak=10\end{cases}}\)
+) Với \(kak=-2\)thì \(\hept{\begin{cases}x=4kak=-8\\y=3kak=-6\\z=5kak=-10\end{cases}}\)
Vậy ...
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Ta có : \(xyz=-30\)
\(\Leftrightarrow2k\times3k\times5k=-30\)
\(\Leftrightarrow30k^3=-30\)
\(\Leftrightarrow k^3=-1\)
\(\Leftrightarrow k=-1\)
Thay vào ta được :
\(\hept{\begin{cases}x=2k=-2\\y=3k=-3\\z=5k=-5\end{cases}}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{x}{3}=\frac{y}{4};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)
Theo tính chất dãy tỉ số bằng nhau
\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\Rightarrow x=27;y=36;z=60\)
b, \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\)
Theo tính chất dãy tỉ số bằng nhau
\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)
\(\Rightarrow x=18;y=24;z=30\)
c, \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-4}{4}\)
Theo tính chất dãy tỉ số bằng nhau
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-4}{4}=\frac{2x+3y-z-2-6+4}{4+9-4}=\frac{46}{9}\)
\(\Rightarrow x=\frac{101}{9};y=\frac{52}{3};z=\frac{220}{9}\)
d, Đặt \(x=2k;y=3k;z=5k\Rightarrow xyz=810\Rightarrow30k^3=810\)
\(\Leftrightarrow k^3=27\Leftrightarrow k=3\)Với k = 3 thì \(x=6;y=9;z=15\)
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Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k,y=3k,z=5k\)
Ta có:
\(xyz=810\\ \Rightarrow2k.3k.5k=810\\ \Rightarrow30k^3=810\\ \Rightarrow k^3=810:30\\ \Rightarrow k^3=27\\ \Rightarrow k=3\)
Vậy:
x = 2k = 2.3 = 6
y = 3k = 3.3 = 9
z = 5k = 5.3 = 15
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B1 :
\(\frac{x}{3}=\frac{y}{6}=\frac{xy}{3\times6}=\frac{162}{18}=9\)
---> x = 3.9 = 27
---> y = 6.9 = 54
B2 :
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{xyz}{2\times3\times5}=\frac{-240}{30}=-8\)
---> x = -8.2 = -16
---> y = -8.3 = -24
---> z = -8.5 = -40
xin tiick
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Ta có:\(\frac{4}{x+1}=\frac{2}{y-2}=\frac{3}{z+2}\)\(\Rightarrow\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}\)
Đặt \(\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}=k\)
\(\Rightarrow x=4k-1,y=2k+2,z=3k-2\)
Theo đề ta có:xyz=12
\(\Rightarrow\left(4k-1\right)\left(2k+2\right)\left(3k-2\right)=12\)
\(\Rightarrow\left(8k^2+8k-2k-2\right)\left(3k-2\right)=12\)
\(\Rightarrow\left(8k^2+6k-2\right)\left(3k-2\right)=12\)
\(\Rightarrow\left(8k^2+6k\right)\left(3k-2\right)-2\left(3k-2\right)\)
\(\Rightarrow24k^3-16k^2+18k^2-12k-6k+4=12\)
\(\Rightarrow24k^3+2k^2-18k=8\)
\(\Rightarrow24k^3+2k^2-18k-8=0\)
\(\Rightarrow\left(k-1\right)\left(24k^2+26k+8\right)=0\)(làm hơi tắt)
TH1:k-1=0,k=1
TH2:\(\left(24k^2+26k+8\right)=0\)
\(24\left(k+\frac{13}{24}\right)^2+\frac{23}{24}>0\)(vô lí)
\(\Rightarrow k=1\)
\(\Rightarrow x=3,y=4,z=1\)
Đặt \(\frac{x}{5}=\frac{y}{2}=\frac{z}{-3}=k\) => x = 5k ; y = 2k ; z = -3k
Ta có :
xyz = 5k . 2k . (-3k) = -30k3 = 240
=> k3 = 240 : (-30) = -8 = -23
=> k = -2
Do đó x = -2 . 5 = -10 ;
y = -2 . 2 = -4 ;
z = -2 . (-3) = 6
Quả nhiên nói cục nam châm hút l-i-k-e là không sai