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Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\). Khi đó ta có:
\(x=2k;2y=3k\Rightarrow y=\dfrac{3k}{2};3z=4k\Rightarrow z=\dfrac{4k}{3}\)
\(\Rightarrow xyz=108\Leftrightarrow2k\cdot\dfrac{3k}{2}\cdot\dfrac{4k}{3}=108\)
\(\Rightarrow\dfrac{24k^3}{6}=108\Rightarrow k^3=27\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=\dfrac{3\cdot3}{2}=\dfrac{9}{2}\\z=\dfrac{4\cdot3}{3}=4\end{matrix}\right.\)
Vậy....
Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\2y=3k\Rightarrow y=\dfrac{3k}{2}\\3z=4k\Rightarrow z=\dfrac{4k}{3}\end{matrix}\right.\)
Mà \(xyz=108\)
\(\Rightarrow2k.\dfrac{3k}{2}.\dfrac{4k}{3}=108\)
\(\Rightarrow2k.\dfrac{3}{2}k.\dfrac{4}{3}k=108\)
\(\Rightarrow k^3.4=108\)
\(\Rightarrow k^3=\dfrac{108}{4}=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=\dfrac{3.3}{2}=4,5\\z=\dfrac{4.3}{3}=4\end{matrix}\right.\)
Vậy \(x=6;y=4,5;z=4\)
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
\(\Leftrightarrow\hept{\begin{cases}x=3k=3.10=30\\y=\frac{2}{k}=\frac{2}{10}=\frac{1}{5}\\z=4k=4.10=40\end{cases}}\)
bt bài này là tỉ lệ thức nhưng sau đợt nghỉ tớ vẫn nhớ đc xương xương :v
\(+,\frac{x}{y}=\frac{10}{9}=\frac{x}{10}=\frac{y}{9}\) (1)
\(+,\frac{y}{z}=\frac{3}{4}=\frac{y}{3}=\frac{z}{4}\)( 2 )
đến đây tại s nhể quên mất òi
Từ (1) ; (2) \(\Rightarrow\frac{x}{10}=\frac{y}{9};\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x}{10}=\frac{3y}{27};\frac{9y}{27}=\frac{z}{4}\Rightarrow\frac{x}{30}=\frac{y}{27}=\frac{z}{36}\)
ADTC dãy tỉ số bằng nhau ta cs
\(\frac{x}{30}=\frac{y}{27}=\frac{z}{36}=\frac{x-2y+3z}{30-2.27+3.36}=\frac{168}{84}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{30}=2\\\frac{y}{27}=2\\\frac{z}{36}=2\end{cases}\Rightarrow\hept{\begin{cases}x=2.30=60\\y=2.27=54\\z=2.36=72\end{cases}}}\)
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Bài làm
Ta có \(\frac{x}{y}=\frac{10}{9};\frac{y}{z}=\frac{3}{4}\)
\(\Leftrightarrow\frac{x}{10}=\frac{y}{9};\frac{y}{3}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{10}=\frac{y}{9};\frac{y}{9}=\frac{z}{12}\)
\(\Leftrightarrow\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\)
Đặt \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=k\)
\(\Leftrightarrow\hept{\begin{cases}x=10k\\y=9k\\z=12k\end{cases}}\)
Thay x = 10k ; y = 9k ; z = 12k vào x - 2y + 3x = 168 ta có
10k - 2.9k + 3.12k = 168
<=> 10k - 18k + 36k = 168
<=> k ( 10 - 18 + 36 ) = 168
<=> k . 28 = 168
<=> k = 6
\(\Leftrightarrow\hept{\begin{cases}x=10.6=60\\y=9.6=54\\z=12.6=72\end{cases}}\)
Vậy x = 60; y = 54 và z = 72
@@ Học tốt
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=>\frac{x-1}{2}=\frac{2\left(y-2\right)}{6}=\frac{3\left(z-3\right)}{12}=>\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)
Theo t/c dãy tỉ số=nhau:
\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{x-1-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}=\frac{x-1-2y+4+3z-9}{8}\)
\(=\frac{\left(x-2y+3z\right)-\left(1-4+9\right)}{8}=\frac{14-6}{8}=\frac{8}{8}=1\)
Do đó: \(\frac{x-1}{2}=1=>x-1=2=>x=3\)
\(\frac{y-2}{3}=1=>y-2=3=>y=5\)
\(\frac{z-3}{4}=1=>z-3=4=>z=7\)
Vậy x=3;y=5;z=7
\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chát dãy tỉ số = nhau ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
\(\Rightarrow\frac{x}{10}=2\Rightarrow x=20\)
\(\frac{y}{15}=2\Rightarrow y=30\)
\(\frac{z}{21}=3\Rightarrow z=63\)
b, Tự làm
c, \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)
\(2x=3z\Leftrightarrow\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{5};\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{x}{6}=\frac{z}{10}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)
Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k(k\inℤ)\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\)
\(\Leftrightarrow x\cdot y=6k\cdot15k=90\)
\(\Leftrightarrow90:k^2=90\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=15\\z=10\end{cases}}\)hay \(\hept{\begin{cases}x=-6\\y=-15\\z=-10\end{cases}}\)
Vậy \((x,y)\in(6,15);(-6,-15)\)
suy ra
6xyz / 24 = xyz / 4 = 108/4 = 27
x=54
y=81/2
z=36
Nhân các vế lại với nhau :
=>6xyz / 24 = xyz / 4 = 108/4 = 27
x=54
y=81/2
z = 27x4:3=36
Vậy .................