\(x\cdot y=20\)\(\Rightarrow x=\frac{20}{y}\)

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:\(\frac{x}{5}=\frac{y}{4}=\frac{x.y}{5.4}=\frac{20}{20}=1\)

*\(\frac{x}{5}=1\Rightarrow x=5;\frac{y}{4}=1\Rightarrow y=4\)

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 \(k=-2\Rightarrow\hept{\begin{cases}x=4k=4\cdot\left(-2\right)=-8\\y=9k=9\cdot\left(-2\right)=-18\end{cases}}\)

vì x/5=y/4=> 4x=5y mà x.y=20=>

x=5,y=4

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

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)}

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

\(\frac{x}{3}=\frac{y}{7}\) 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\(\Rightarrow x=-4.3=-12;y=-4.7=-28\)

x/3=y/7=x.y/3.7=84/21=4

=>x=3.4=12

=>y=7.4=28

bạn khỏi bam đug cho mìh nhé

\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x}.\frac{1}{y}\)

\(=>\frac{y-x}{xy}=\frac{1}{xy}\)

\(=>xy^2-x^2y=xy\)

\(=>xy^2-x^2y-xy=0\)

\(=>x.\left(y^2-xy-y\right)=0\)

\(=>\orbr{\begin{cases}x=0\\y^2-xy-y=0\end{cases}}\)

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 \(y\ne1;x\ne-1;x\ne0;y\ne0\)

Ta có\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x}.\frac{1}{y}\)

\(\Leftrightarrow\frac{y-x}{xy}=\frac{1}{xy}\)

\(\Leftrightarrow y-x=1\Rightarrow y=x+1\)

Vậy..................

Ko có giá trị cụ thể nha

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 \(x=\pm6\)

\(y=\pm14\)

\(\frac{x}{3}=\frac{y}{7}\)

\(\Rightarrow x=3;y=7\)

\(x.y=84\)

\(\Rightarrow x=\frac{84}{y}\)

\(x-0,27=\frac{\frac{73}{77}+\frac{73}{165}+\frac{73}{285}}{28.\left(\frac{5}{84}+\frac{3}{285}+\frac{4}{285}\right)}\)

\(x-\frac{27}{100}=\frac{73.\left(\frac{1}{77}+\frac{1}{165}+\frac{1}{285}\right)}{25.4.\left(\frac{5}{4.84}+\frac{3}{4.180}+\frac{4}{4.285}\right)}\)

\(x-\frac{27}{100}=\frac{73,\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{5}{366}+\frac{3}{240}+\frac{4}{285}\right)}\)

\(x-\frac{27}{100}=\frac{73.\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{2}{105}+\frac{1}{285}\right)}\)

\(x-\frac{27}{100}=\frac{73}{100}\)

\(x=\frac{27}{100}+\frac{73}{100}\)

\(\Rightarrow x=1\)

bạn nguyen thi thuy linh ơi bạn làm sai đề của mk rồi