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Ta có :\(15x=10y=6z\Rightarrow\hept{\begin{cases}15x=10y\\10y=6z\end{cases}}\Rightarrow\hept{\begin{cases}3x=2y\\5y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{5}\end{cases}}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Khi đó 5x3 + 2y3 - z3 = 31
=> 5(2k)3 + 2(3k)3 - (5k)3 = 31
=> 40k3 + 54k3 - 125k3 = 31
=> -31k3 = 31
=> k3 = -1
=> k = -1
=> x = -2 ; y = -3 ; z = -5
b) Ta có 7x = 14y = 6z => \(\hept{\begin{cases}7x=14y\\14y=6z\end{cases}}\Rightarrow\hept{\begin{cases}x=2y\\7y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{1}\\\frac{y}{3}=\frac{z}{7}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{6}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{7}\end{cases}}\Rightarrow\frac{x}{6}=\frac{y}{3}=\frac{z}{7}\)
Đặt \(\frac{x}{6}=\frac{y}{3}=\frac{z}{7}=k\Rightarrow\hept{\begin{cases}x=6k\\y=3k\\z=7k\end{cases}}\)
Khi đó 2x2 - 3y2 = 5
<=> 2.(6k)2 - 3.(3k)2 = 5
=> 72k2 - 27k2 = 5
=> 45k2 = 5
=> k2 = 1/9
=> k = \(\pm\frac{1}{3}\)
Nếu k = 1/3 => x = 2 ; y = 1 ; z = 7/3
Nếu k = -1/3 => x = -2 ; y = - 1 ; z = -7/3
Vậy các cặp (x;y;z) thỏa mãn là : (2;1;7/3) ; (-2 ; - 1; -7/3)
c) Ta có : \(3x=8y=5z\Rightarrow\frac{3x}{120}=\frac{8y}{120}=\frac{5z}{120}\Rightarrow\frac{x}{40}=\frac{y}{15}=\frac{z}{24}\)
Đặt \(\frac{x}{40}=\frac{y}{15}=\frac{z}{24}=k\Rightarrow\hept{\begin{cases}x=40k\\y=15k\\z=24k\end{cases}}\)
Khi đó |x - 2y| = 5
<=> |40k - 2.15k| = 5
=> |10k| = 5
=> \(\orbr{\begin{cases}10k=5\\10k=-5\end{cases}}\Rightarrow\orbr{\begin{cases}k=\frac{1}{2}\\k=-\frac{1}{2}\end{cases}}\)
Nếu k = 5 => x = 20 ; y = 7,5 ; z = 12
Nếu k = -5 => x = -20 ; y =-7,5 ; z = -12
d) 4x = 5y = 6z => \(\frac{4x}{60}=\frac{5y}{60}=\frac{6z}{60}\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{10}\)
Đặt \(\frac{x}{15}=\frac{y}{12}=\frac{z}{10}=k\Rightarrow\hept{\begin{cases}x=15k\\y=12k\\z=10k\end{cases}}\)
Khi đó (3x - 2y)2 = 16
<=> (3.15k - 2.12k)2 = 16
=> (45k -24k)2 = 16
=> (21k)2 = 16
=> \(\orbr{\begin{cases}21k=4\\21k=-4\end{cases}}\Rightarrow\orbr{\begin{cases}k=\frac{4}{21}\\k=-\frac{4}{21}\end{cases}}\)
Nếu k = 4/21 => x = 20/7 ; y = 16/7 ; z = 40/21
Nếu k = -4/21 => x = -20/7 ; y = -16/7 ; z = -40/21
Đặt \(\frac{x}{-5}=\frac{y}{6}=\frac{z}{-2}=k\) \(\left(k\ne0\right)\)
\(\Rightarrow x=-5k;y=6k;z=-2k\)
\(\Rightarrow A=\frac{3.k.\left(-5\right)+6.k-2.\left(-2\right).k}{-3.\left(-5\right).k-5.6.k+6.\left(-2\right).k}=\frac{-15k+6k+4k}{15k-30k-12k}=\frac{-5k}{-27k}=\frac{5}{27}\)
Vậy \(A=\frac{5}{27}\).
Vì 3x = 5y = 6z
=> \(\frac{x}{5}=\frac{y}{3};\frac{y}{6}=\frac{z}{5}\)
\(=>\frac{x}{30}=\frac{y}{18};\frac{y}{18}=\frac{z}{15}\)
\(hay\)\(\frac{x}{30}=\frac{y}{18}=\frac{z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{30}=\frac{y}{18}=\frac{z}{15}=\frac{x+y-z}{30+18-15}=\frac{22}{33}=\frac{2}{3}\)
Do đó suy ra:
\(3x=\frac{2}{3}=>x=\frac{2}{9}\)
\(5y=\frac{2}{3}=>y=\frac{2}{15}\)
\(6z=\frac{2}{3}=>x=\frac{1}{9}\)
Vậy \(\left(x;y;z\right)\in\left\{\frac{2}{9};\frac{2}{15};\frac{1}{9}\right\}\)
\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)
\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)
Ta có: 3x = 4y => \(\frac{x}{4}=\frac{y}{3}\) => \(\frac{x}{8}=\frac{y}{6}\)
5y = 6z => \(\frac{y}{6}=\frac{z}{5}\)
=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y-z}{8+6-5}=\frac{18}{9}=2\)
=> \(\hept{\begin{cases}\frac{x}{8}=2\\\frac{y}{6}=2\\\frac{z}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.8=16\\y=2.6=12\\z=2.5=10\end{cases}}\)
Vậy ....
3x=4y
=>x/4=y/3
=>x/8=y/6
5y=6z
=>y/6=z/5
=>x/8=y/6=z/5
Đặt x/8=y/6=z/5=k
=>x=8k; y=6k; z=5k
xyz=30
=>8k*6k*5k=30
=>240k^3=30
=>k^3=1/8
=>k=1/2
=>x=8*1/2=4; y=6*1/2=3; z=5*1/2=5/2
\(\frac{x}{5}=\frac{y}{3},\frac{y}{6}=\frac{z}{21}\)
\(\frac{x}{30}=\frac{y}{18},\frac{y}{18}=\frac{z}{63}\)
\(\Rightarrow\frac{x}{30}=\frac{y}{18}=\frac{z}{63}=\frac{5x}{150}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x}{30}=\frac{y}{18}=\frac{z}{63}=\frac{5x}{150}=\frac{5x+y-z}{150+18-63}=\frac{28}{105}=\frac{4}{15}\)
đến đây tự tính tiếp