Question

x/-3 = y/7 và x^2 -y^2 = -160

Answer 1

Đặt \(\frac{x}{-3}=\frac{y}{7}=k\)

=>x=-3k; y=7k

\(x^2-y^2=-160\)

=>\(\left(-3k\right)^2-\left(7k\right)^2=-160\)

=>\(9k^2-49k^2=-160\)

=>\(-40k^2=-160\)

=>\(k^2=4\)

=>\(\left[\begin{array}{l}k=2\\ k=-2\end{array}\right.\)

TH1: k=2

=>\(\begin{cases}x=-3\cdot2=-6\\ y=7\cdot2=14\end{cases}\)

TH2: k=-2

=>\(\begin{cases}x=\left(-3\right)\cdot\left(-2\right)=6\\ y=7\cdot\left(-2\right)=-14\end{cases}\)

Answer 2

Đặt\(\frac{x}{-3}=\frac{y}{-7}=k\)
=> \(x=-3k\), \(y=7k\)

=> \(\left(-3k^{}\right)^2-\left(7k\right)^2=-160\)

=> \(9k^2-49k^2=-160\)

=> \(k^2\left(9-49\right)=-160\)

=> \(-40k^2=-160\)

=> \(k^2=\frac{-160}{-40}=4\)

=> \(k^2=4\)

=>\(k^2=\left\lbrace2;-2\right\rbrace\)

Khi k = 2 => x = -3.2 = -6 ; y = 7.2 = 14

Khi k = -2 => x= -3.-2 = 6; y = 7.-2 = -14

Vậy \(\left\lbrace x;y\right\rbrace\in\left\lbrace\left(-6;14\right);\left(6;-14\right)\right\rbrace\)

Answer 3

(x;y)∈{(−6;14);(6;−14)} nhé, mik nhầm
chuk bạn hok tốt nhaa

Answer 4

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