\(a.\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\&xy=1200\)
\(\Leftrightarrow\dfrac{15}{20}=\dfrac{x-9}{y-12}\Leftrightarrow\dfrac{3}{4}=\dfrac{x-9}{y-12}\)
\(\Rightarrow\dfrac{9}{12}=\dfrac{x-9}{y-12}\)
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 \(\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}=k\Rightarrow x=-20k;y=30k;z=-50k\)
\(\Rightarrow xyz=30000\Rightarrow-20k.30k.\left(-50k\right)=30000\Rightarrow30000k^3=30000\)
\(\Rightarrow k^3=1\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=30\\z=-50\end{matrix}\right.\)
⇔1520=x−9y−12⇔34=x−9y−12⇔1520=x−9y−12⇔34=x−9y−12
912=x−9y−12=x−9+9y−12+12=xy912=x−9y−12=x−9+9y−12+12=xy
34=xyy2⇒34=1200y2⇒y2=1200.43=160034=xyy2⇒34=1200y2⇒y2=1200.43=1600
⇒y=√1600=±40⇒y=1600=±40
+ TH1: y=40⇒x=30y=40⇒x=30
15x−9=40z−4⇒z=−8015x−9=40z−4⇒z=−80
Vậy, các cặp (x;y;z)(x;y;z) thỏa mãn là (30;40;80)&(−30;−40;−80)