Question

cho \(\frac{x}{5}\) =\(\frac{y}{3}\) và \(^{x^2}\) -\(y^2\) =4

Answer 1

Giải:

Đặt \(\frac{x}{5}=\frac{y}{3}=k\)

\(\Rightarrow x=5k,y=3k\)

Mà \(x^2-y^2=4\)

\(\Rightarrow\left(5k\right)^2-\left(3k\right)^2=4\)

\(\Rightarrow5^2.k^2-3^2.k^2=4\)

\(\Rightarrow k^2\left(5^2-3^2\right)=4\)

\(\Rightarrow k^2.16=4\)

\(\Rightarrow k^2=\frac{1}{4}\)

\(\Rightarrow k=\pm\frac{1}{2}\)

+) \(k=\frac{1}{2}\Rightarrow x=\frac{5}{2},y=\frac{3}{2}\)

+) \(k=\frac{-1}{2}\Rightarrow x=\frac{-5}{2},y=\frac{-3}{2}\)

Vậy cặp số \(\left(x;y\right)\) là \(\left(\frac{5}{2};\frac{3}{2}\right);\left(\frac{-5}{2};\frac{-3}{2}\right)\)

Answer 2

Theo bài ra ta có: \(\frac{x}{5}=\frac{y}{3}\)

=> \(\frac{x^2}{25}=\frac{y^2}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{4}{16}=\frac{1}{4}\)

=> x²=\(\frac{1}{4}.25=\frac{25}{4}\)=>x=\(\pm\frac{5}{2}\)

y²=\(\frac{1}{4}.9=\frac{9}{4}\)=>y=\(\pm\frac{3}{2}\)

Vì \(\frac{x}{5}=\frac{y}{3}\)=> x và y cùng dấu

Vậy (x;y) thõa mãn là (\(\frac{5}{2};\frac{3}{2}\)); \(\left(\frac{-5}{2};\frac{-3}{2}\right)\)

Answer 3

Ta có:\(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)

\(\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{4}{16}=\frac{1}{4}\)(T/C...)

\(\Rightarrow x^2=\frac{1}{4}\cdot25=\frac{25}{4}\Rightarrow x=\pm\frac{5}{2}\)

\(\Rightarrow y^2=\frac{1}{4}\cdot9=\frac{9}{4}\Rightarrow y=\pm\frac{3}{2}\)

Vậy \(x=\frac{5}{2},y=\frac{3}{2}\) hoặc \(x=-\frac{5}{2},y=-\frac{3}{2}\)

Answer 4

\(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{5^2}=\frac{y^2}{3^2}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{4}{16}=\frac{1}{4}\)

=> x = 5.1/4 = 5/4

y = 3.1/4 = 3/4