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40/x-30=20/y-15=28/z-21 => 40/x-40/30=20/y-20/15=28/z-28/21 => 40/x-4/3=20/y-4/3=28/z-4/3
<=> 40/x=20/y=28/z=K => x=40.K; y=20.K; z=28.K
<=> xyz=40.20.28.K3 => xyz=22400.K3
<=>K3=1 => K=+-1
<=> x=40.K = 40.1=40 (1)
=40.(-1)=-40
TH(1): x=40 => y=20; z =28
TH(2); x=-40 => y=-20; z=-28
vậy x=40; y=20; z =28
hoặc x=-40; y=-20; z=-28
câu b làm y vậy đó bạn đổi 15x=-10y=6z=>x/1/15=y/-1/10=z/1/6
\(\Rightarrow\left[\begin{array}{nghiempt}x-9=15k\\y-12=20k\\z-24=40k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=15k+9\\y=20k+12\\z=40k+24\end{array}\right.}\)
ta có:
x.y=1200\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\Rightarrow\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}=k\)
=> (15k+9)(20k+12)=1200
=> 3.4(5k+3)(5k+3)=1200
=> (5k+3)2=100
=> 5k+3=\(\pm\)10
=> \(\left[\begin{array}{nghiempt}5k+3=10\\5k+3=-10\end{cases}\Rightarrow\left[\begin{array}{nghiempt}5k=7\\5k=-13\end{cases}\Rightarrow}\left[\begin{array}{nghiempt}k=\frac{7}{5}\\k=-\frac{13}{5}\end{array}\right.}\)
* với k=7/5
x=7/5x15+9=30
y=7/5x20+12=40
z=7/5x40+24=80
* với k=-13/5
x=-13/5x15+9=-30
y=-13/5x20+12=-40
z=-13/5x40+24=-80
b)
\(\frac{40}{x-30}=\frac{20}{y-50}=\frac{28}{z-21}\Rightarrow\frac{x-30}{40}=\frac{y-50}{20}=\frac{z-21}{28}k=\)
=>\(\left[\begin{array}{nghiempt}x-30=40k\\y-50=20k\\z-21=28k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=40k+30\\y=20k+50\\z=28k+21\end{array}\right.}\)
ta có:
x.y.z=22400
=> (40k+30)(20k+50)(28k+21)=22400
c) 15x=-10y=6z
\(\Rightarrow\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\Rightarrow\frac{x}{2}=-\frac{y}{3}=\frac{z}{5}=k\)
=> \(\left[\begin{array}{nghiempt}x=2k\\y=-3k\\z=5k\end{array}\right.\)
ta có:
x.y.z=30000
=> 2k.(-3k).5k=30000
=> k3=1000
=> k=10
ta có: x=10x2=20
y=10.(-3)=-30
z=10.5=50
\(a.\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\&xy=1200\)
\(\Leftrightarrow\dfrac{15}{20}=\dfrac{x-9}{y-12}\Leftrightarrow\dfrac{3}{4}=\dfrac{x-9}{y-12}\)
\(\Rightarrow\dfrac{9}{12}=\dfrac{x-9}{y-12}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{9}{12}=\dfrac{x-9}{y-12}=\dfrac{x-9+9}{y-12+12}=\dfrac{x}{y}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{xy}{y^2}=\dfrac{x^2}{xy}\)
Từ \(\dfrac{3}{4}=\dfrac{xy}{y^{^2}}\Rightarrow\dfrac{3}{4}=\dfrac{1200}{y^2}\Rightarrow y^2=1200.\dfrac{4}{3}=1600\)
\(\Rightarrow y=\sqrt{1600}=\pm40\)
+ TH1: \(y=40\Rightarrow x=30\)
\(\dfrac{15}{x-9}=\dfrac{40}{z-24}\Rightarrow z=80\) (tự giải pt)
+ TH2: \(y=-40\Rightarrow x=-30\)
\(\dfrac{15}{x-9}=\dfrac{40}{z-4}\Rightarrow z=-80\) (tự giải pt)
Vậy, các cặp \(\left(x;y;z\right)\) thỏa mãn là \(\left(30;40;80\right)\&\left(-30;-40;-80\right)\)
\(b.15x=-10y=6z\&xyz=30000\)
\(\Rightarrow\left\{{}\begin{matrix}15x=-10y\\-10y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-10}=\dfrac{y}{15}\\\dfrac{y}{6}=\dfrac{z}{-10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-20}=\dfrac{y}{30}\\\dfrac{y}{30}=\dfrac{z}{-50}\end{matrix}\right.\Rightarrow\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}\)
Đặt \(\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}=k\Rightarrow x=-20k;y=30k;z=-50k\)
\(\Rightarrow xyz=30000\Rightarrow-20k.30k.\left(-50k\right)=30000\Rightarrow30000k^3=30000\)
\(\Rightarrow k^3=1\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=30\\z=-50\end{matrix}\right.\)
a.\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\)
=>\(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\)
=>\(\frac{x}{15}-\frac{9}{15}=\frac{y}{20}-\frac{12}{20}=\frac{z}{40}-\frac{24}{40}\)
=>\(\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)
=>\(\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\)
Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\Rightarrow x=15k,y=20k,z=40k\)
Ta có: \(xy=15k.20k=300k^2=1200\Rightarrow k^2=4\Rightarrow k=\pm2\)
Với k = 2 => x = 30, y = 40, z = 80
Với k = -2 => x=-30,y=-40,z=-80
Vậy...
b tương tự a
c, \(15x=-10y=6z\Rightarrow\frac{x}{\frac{1}{15}}=\frac{y}{\frac{-1}{10}}=\frac{z}{\frac{1}{6}}=k\Rightarrow x=\frac{1}{15}k,y=\frac{-1}{10}k,z=\frac{1}{6}k\)
Ta có: \(xyz=\frac{1}{15}k\cdot\frac{-1}{10}k\cdot\frac{1}{6}k=\frac{-1}{900}k^3=-30000\Rightarrow k^3=27000000\Rightarrow k=300\)
=> x = 20, y = -30, z = 50