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\(x\cdot y=20\)\(\Rightarrow x=\frac{20}{y}\)
thay vào biểu thức ta đc
\(\frac{\frac{20}{y}}{5}=\frac{y}{4}\)
\(\frac{20}{y}\cdot4=5y\)
\(\frac{80}{y}=5y\)
\(80=5y^2\)
\(16=y^2\)
\(\Rightarrow y=4\)
\(\Rightarrow x=\frac{20}{4}=5\)
hok tốt
\(\frac{x}{5}=\frac{y}{4}\) biết x.y=20
Ta có:\(\frac{x}{5}=\frac{y}{4}=\frac{x.y}{5.4}=\frac{20}{20}=1\)
*\(\frac{x}{5}=1\Rightarrow x=5;\frac{y}{4}=1\Rightarrow y=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{x}{4}=\frac{y}{9}=k\Rightarrow x=4k;y=9k\)
\(\Rightarrow xy=144\Leftrightarrow4k\cdot9k=144\)
\(\Rightarrow36k^2=144\)
\(\Rightarrow k^2=4\Rightarrow k=\pm2\)
Nếu \(k=2\Rightarrow\hept{\begin{cases}x=4k=4\cdot2=8\\y=9k=9\cdot2=18\end{cases}}\)
Nếu \(k=-2\Rightarrow\hept{\begin{cases}x=4k=4\cdot\left(-2\right)=-8\\y=9k=9\cdot\left(-2\right)=-18\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt
x/5=y/4=k
khi đó:
x=5k
y=4k
Ta lại có:
x.y=4k.5k=20k^2=20
=> K=+-1
Khi k=1
Khi k=-1
Giải ra nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt x/3=y/=k(k khác 0) =>x=3k;y=7k
=>x.y=3k.7k=21.k^2=84
=>k^2=4=(2)^2 hoặc(-2)^2
th1:k=2=> x=6;y=14
th2:k=-2 =>x=-6;y=-14
Đặt \(\frac{x}{3}=\frac{y}{7}=k\) ta có :
\(x=3k\) ;\(y=7k\)
Vì \(x.y=84\Rightarrow3k.7k=21k^2=84\)
\(\Rightarrow k^2=4=2^2\)
\(\Rightarrow\orbr{\begin{cases}k=-2\\k=2\end{cases}}\)
+TH1: \(k=-2\Rightarrow\hept{\begin{cases}x=-6\\y=-14\end{cases}}\)
+TH2: \(k=2\Rightarrow\hept{\begin{cases}x=6\\y=14\end{cases}}\)
Vậy (x,y) = {(-6,-14);(6,14)}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{3}=\frac{y}{7}\) va xy=84
Dat : \(\frac{x}{3}=\frac{y}{7}=k\)
x.y=21k2
84 =21k2
k2 = 4
k = +-2
Neu : k=4\(\Rightarrow x=4.3=12;y=4.7=28\)
Neu : k=-4\(\Rightarrow x=-4.3=-12;y=-4.7=-28\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x}.\frac{1}{y}\)
\(=>\frac{y-x}{xy}=\frac{1}{xy}\)
\(=>xy^2-x^2y=xy\)
\(=>xy^2-x^2y-xy=0\)
\(=>x.\left(y^2-xy-y\right)=0\)
\(=>\orbr{\begin{cases}x=0\\y^2-xy-y=0\end{cases}}\)
Ta thấy \(y^2-xy-y=0\)
\(=>y.\left(y-x-y\right)=0\)
\(=>\orbr{\begin{cases}y=0\left(2\right)\\y-y=0\end{cases}}\)
Từ 1 và 2 => x = y = 0
\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x}.\frac{1}{y}\)
\(\Rightarrow\frac{y-x}{xy}=\frac{1}{xy}\)
\(\Rightarrow y-x=1\)
Vậy x,y có dạng \(\hept{\begin{cases}x=y-1\\y=x+1\end{cases}}\)với \(y\ne1;x\ne-1;x\ne0;y\ne0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: xy = 84
=> \(y=\frac{84}{x}\)
=> \(\frac{x}{3}=\frac{\frac{84}{x}}{7}\)
=> \(\frac{x}{3}=\frac{12}{x}\)
=> \(x^2=3.12=36\)
=> \(x=\pm6\)
Khi x = 6
=> \(y=\frac{84}{x}=\frac{84}{6}=14\)
Khi x = -6
=> \(y=\frac{84}{x}=\frac{84}{-6}=-14\)
Theo bài ra ta có: \(\frac{x}{3}=\frac{y}{7}\Rightarrow\frac{x}{3}.\frac{x}{3}=\frac{y}{7}.\frac{y}{7}=\frac{x}{3}.\frac{y}{7}\)
\(\Rightarrow\frac{x^2}{9}=\frac{y^2}{49}=\frac{84}{21}=4\)
\(\Rightarrow x^2=4.9=36\Rightarrow x=\pm6\)
\(\Rightarrow y^2=196=\pm14\)
Vậy \(x=\pm6\)
\(y=\pm14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: \(\frac{3}{x}=\frac{y}{7}=\frac{x}{3}=\frac{y}{7}\)
Đặt \(\frac{x}{3}=\frac{y}{7}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)
Thay \(x=3k;y=7k\) vào \(x.y=84\), ta có:
\(3k.7k=84\\ \Leftrightarrow21k^2=84\\ \Leftrightarrow k^2=4\\ \Leftrightarrow k^2=\left(\pm2\right)^2\\ \Rightarrow k\in\left\{2;-2\right\}\)
+Khi \(k=2\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.7=14\end{matrix}\right.\)
+Khi \(k=-2\Rightarrow\left\{{}\begin{matrix}x=-2.3=-6\\y=-2.7=-14\end{matrix}\right.\)
Vậy...
Ta có: \(\frac{3}{x}=\frac{y}{7}.\)
\(\Rightarrow\frac{x}{3}=\frac{y}{7}\) và \(x.y=84.\)
Đặt \(\frac{x}{3}=\frac{y}{7}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)
Có: \(x.y=84\)
=> \(3k.7k=84\)
=> \(21k^2=84\)
=> \(k^2=84:21\)
=> \(k^2=4\)
=> \(k=\pm2.\)
TH1: \(k=2.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.7=14\end{matrix}\right.\)
TH2: \(k=-2.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-2\right).3=-6\\y=\left(-2\right).7=-14\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;14\right),\left(-6;-14\right).\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
Do x/2 = y/3 => 3x = 2y
=> x = 2/3y
Ta có: x.y = 54
=> 2/3y.y = 54
=> y2 = 54 : 2/3
=> y2 = 54 . 3/2
=> y2 = 81
=> y thuộc {9 ; -9}
+ Với y = 9 => x = 2/3.9 = 6
+ Với y = -9 => x = 2/3.(-9) = -6
Vậy x = 6; y = 9 hoặc x = -6; y = -9
Ta có:
\(\frac{x}{4}=\frac{y}{7}\) và \(x.y=84.\)
Đặt \(\frac{x}{4}=\frac{y}{7}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)
Có: \(x.y=84\)
=> \(4k.7k=84\)
=> \(28.k^2=84\)
=> \(k^2=84:28\)
=> \(k^2=3\)
=> \(k^2=\left(\pm\sqrt{3}\right)^2\)
=> \(k=\pm\sqrt{3}.\)
TH1: \(k=\sqrt{3}.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\sqrt{3}=4\sqrt{3}\\y=7.\sqrt{3}=7\sqrt{3}\end{matrix}\right.\)
TH2: \(k=-\sqrt{3}.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-\sqrt{3}\right)=4\left(-\sqrt{3}\right)\\y=7.\left(-\sqrt{3}\right)=7\left(-\sqrt{3}\right)\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(4\sqrt{3};7\sqrt{3}\right);\left[4\left(-\sqrt{3}\right);7\left(-\sqrt{3}\right)\right].\)
Chúc bạn học tốt!
help me