Question

\(\dfrac{x}{5}\)=\(\dfrac{y}{7}\)=\(\dfrac{z}{3}\) và x² +y²-z²=585

Answer 1

Đặt \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=k\)

=>x=5k; y=7k; z=3k

\(x^2+y^2-z^2=585\)

=>\(\left(5k\right)^2+\left(7k\right)^2-\left(3k\right)^2=585\)

=>\(25k^2+49k^2-9k^2=585\)

=>\(65k^2=585\)

=>\(k^2=9\)

=>\(\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)

TH1: k=3

=>\(\left\{{}\begin{matrix}x=5\cdot3=15\\y=7\cdot3=21\\z=3\cdot3=9\end{matrix}\right.\)

TH2: k=-3

=>\(\left\{{}\begin{matrix}x=5\cdot\left(-3\right)=-15\\y=7\cdot\left(-3\right)=-21\\z=3\cdot\left(-3\right)=-9\end{matrix}\right.\)

Answer 2

5x​=7y​=3z​⇒25x2​=49y2​=9z2​

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

x225=y249=z29=x2+y2−z225+49−9=58565=925x2​=49y2​=9z2​=25+49−9x2+y2−z2​=65585​=9

suy ra:

x225=9⇒x2=225⇒x=1525x2​=9⇒x2=225⇒x=15hoặc x=−15x=−15

y249=9⇒y2=441⇒y=2149y2​=9⇒y2=441⇒y=21hoặc y=−21y=−21

z29=9⇒z2=81⇒z=99z2​= 9⇒  z2 =81 ⇒z=9 hoặc z=−9z=−9