Question

[Tô Liên Bạch]

​\(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)​​​và​​​\(x^2+y^2-z^2=585\)​​\(x^2+y^2-z^2=585\)

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Answer 1

[Nguyễn Huy Tú]

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

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}=\frac{x^2+y^2-z^2}{5^2+7^2-3^2}=\frac{585}{65}=9\)

\(x=45;y=63;z=27\)

Answer 2

[Nobi Nobita]

Từ \(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)(1)\(\Rightarrow\left(\frac{x}{5}\right)^2=\left(\frac{y}{7}\right)^2=\left(\frac{z}{3}\right)^2=\frac{x^2}{25}=\frac{y^2}{49}=\frac{z^2}{9}\)

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

\(\frac{x^2}{25}=\frac{y^2}{49}=\frac{z^2}{9}=\frac{x^2+y^2-z^2}{25+49-9}=\frac{585}{65}=9\)

\(\Rightarrow x^2=25.9=225\)\(\Rightarrow x=\pm15\)

\(y^2=49.9=441\)\(\Rightarrow y=\pm21\)

\(z^2=9.9=81\)\(\Rightarrow z=\pm9\)

Từ (1) \(\Rightarrow x,y,z\)phải có cùng dấu âm hoặc dương

Vậy \(\left(x;y;z\right)=\left(-15;-21;-9\right),\left(15;21;9\right)\)