Question

[cát phượng]

có bao nhiêu cặp giá trị (x;y) thỏa mãn: x/-3= y/5 và xy=-5/27

Answer 1

[Aki Tsuki]

Ta có: \(\frac{x}{-3}=\frac{y}{5}\) = \(\frac{xy}{-3y}=\frac{\frac{-5}{27}}{-3y}\)

=> \(\frac{y}{5}=\frac{\frac{-5}{27}}{-3y}\) => \(y.\left(-3\right)y=5.\left(\frac{-5}{27}\right)\)

=> \(y^2.\left(-3\right)=\frac{-25}{27}\) => \(y^2=\frac{-25}{27}:\left(-3\right)=\frac{25}{81}\)

=> \(y=\sqrt{\frac{25}{81}}=\frac{5}{9};y=-\sqrt{\frac{25}{81}}=\frac{-5}{9}\)

+) \(y=\frac{5}{9}\)

=> \(x.\frac{5}{9}=\frac{-5}{27}\)

=> \(x=\frac{-5}{27}:\frac{5}{9}=\frac{-5}{27}.\frac{9}{5}=\frac{-1}{3}\)

+) \(y=\frac{-5}{9}\)

=> \(x.\frac{-5}{9}=\frac{-5}{27}\)

=> \(x=\frac{-5}{27}:\left(\frac{-5}{9}\right)=\frac{-5}{27}.\left(\frac{9}{-5}\right)=\frac{1}{3}\)

Vậy có 2 cặp giá trị (x;y) thỏa mẵn đề bài là:

(x;y) = \(\left(\frac{5}{9};\frac{-1}{3}\right)\)

(x;y) = \(\left(\frac{-5}{9};\frac{1}{3}\right)\)

Answer 2

[cát phượng]

s là -3y v bn?