\(\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\left(\frac{x}{2}\right)\left(\frac{y}{3}\right)=\frac{xy}{6}=\frac{96}{6}=16\)
\(\Rightarrow x^2=64;y^2=144\)
\(\frac{x}{2}=\frac{y}{3}\) nên x , y cùng dấu, từ đó có :
\(\left(x;y\right)\in\left\{\left(8;12\right);\left(-8;-12\right)\right\}\)
Giải:
Ta có: \(\frac{2}{x}=\frac{3}{y}\Rightarrow\frac{x}{2}=\frac{y}{3}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=k\)
\(\Rightarrow x=2k,y=3k\)
Do \(x.y=96\)
\(\Rightarrow2.k.3.k=96\)
\(\Rightarrow k^2.6=96\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=\pm4\)
+) \(k=4\Rightarrow x=8,y=12\)
+) \(k=-4\Rightarrow x=-8,y=-12\)
Vậy các cặp số ( x, y ) là: \(\left(8,12\right);\left(-8,-12\right)\)
\(\frac{2}{x}=\frac{3}{y}\)
=>2y =3x
=>2.\(\frac{96}{x}=3x\)
=>\(\frac{192}{x}=3x\)
=>\(\frac{192}{3}=x^2\)
=>\(x^2=64\)
=>có 2 trường hợp
th1
x=8
th2
x=-8
\(\frac{2}{x}=\frac{3}{y}\Rightarrow\frac{x}{2}=\frac{y}{3}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=k\) \(\Rightarrow x=2k;y=3k\)
Thay \(x=2k;y=3k\) vào \(x.y=96\) , ta có :
\(x.y=96\)
\(\Rightarrow\left(2k\right).\left(3k\right)=96\)
\(\Rightarrow6k^2=96\)
\(\Rightarrow k^2=96:6\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=\pm4\)
\(+\) Nếu \(k=4\Rightarrow x=8;y=12\)
\(+\) Nếu \(k=-4\Rightarrow x=-8;y=-12\)
Vậy các cặp số \(\left(x;y\right)\) là : \(\left(8;12\right);\left(-8;-12\right)\)
