Question

Cho \(\frac{3z-4y}{5}\)=\(\frac{5y-3x}{4}\)=\(\frac{4x-5z}{3}\), x²-z²=36. Hãy tìm x,y,z

Answer 1

Từ giả thiết suy ra : \(\frac{5\left(3z-4y\right)}{25}=\frac{4\left(5y-3x\right)}{16}=\frac{3\left(4x-5z\right)}{9}=\frac{0}{25+16+9}=0\)

( Tính chất dãy tỉ số bằng nhau )

Vì vậy có : \(\left\{{}\begin{matrix}3z-4y=0\\5y-3x=0\\4x-5z=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3z=4y\\5y=3x\\4x=5z\end{matrix}\right.\) \(\Leftrightarrow\frac{z}{4}=\frac{y}{3}=\frac{x}{5}\)

\(\Leftrightarrow\frac{z^2}{16}=\frac{y^2}{9}=\frac{x^2}{25}=\frac{x^2-z^2}{25-16}=\frac{36}{9}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm25\\y=\pm6\\z=\pm8\end{matrix}\right.\)

Answer 2

Ta có: \(\frac{3z-4y}{5}=\frac{5y-3x}{4}=\frac{4x-5z}{3}.\)

\(\Rightarrow\frac{5.\left(3z-4y\right)}{25}=\frac{4.\left(5y-3x\right)}{16}=\frac{3.\left(4x-5z\right)}{9}.\)

\(\Rightarrow\frac{15z-20y}{25}=\frac{20y-12x}{16}=\frac{12x-15z}{9}.\)

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

\(\frac{15z-20y}{25}=\frac{20y-12x}{16}=\frac{12x-15z}{9}=\frac{15z-20y+20y-12x+12x-15z}{25+16+9}=\frac{\left(15z-15z\right)-\left(20y-20y\right)-\left(12x-12x\right)}{50}=\frac{0}{50}=0.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{3z-4y}{5}=0\\\frac{5y-3x}{4}=0\\\frac{4x-5z}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3z-4y=0\\5y-3x=0\\4x-5z=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3z=4y\\5y=3x\\4x=5z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{z}{4}=\frac{y}{3}\\\frac{y}{3}=\frac{x}{5}\\\frac{x}{5}=\frac{z}{4}\end{matrix}\right.\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{z}{4}.\)

\(\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}=\frac{z^2}{16}\) và \(x^2-z^2=36.\)

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

\(\frac{x^2}{25}=\frac{y^2}{9}=\frac{z^2}{16}=\frac{x^2-z^2}{25-16}=\frac{36}{9}=4.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x^2}{25}=4\Rightarrow x^2=100\Rightarrow\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\\\frac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow\left[{}\begin{matrix}y=6\\y=-6\end{matrix}\right.\\\frac{z^2}{16}=4\Rightarrow z^2=64\Rightarrow\left[{}\begin{matrix}z=8\\z=-8\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(10;6;8\right),\left(-10;-6;-8\right).\)

Chúc bạn học tốt!