\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
\(\Rightarrow\frac{x^2}{2^2}=\frac{y^2}{3^2}=\frac{2z^2}{2\cdot4^2}\)
\(\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)
\(\Rightarrow\frac{x^2-y^2+2z^2}{4-9+32}=\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\) mà x2 - y2 + 2z2 = 108
\(\Rightarrow\frac{108}{27}=\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}\)
\(\Rightarrow4=\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}\)
\(\Rightarrow\hept{\begin{cases}x^2=4\cdot4=16\\y^2=9\cdot4=36\\z^2=4\cdot16=64\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\pm4\\y=\pm6\\z=\pm8\end{cases}}\)
vậy_
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}=\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)
Áp dụng tính chất của dãy tỉ số bằng nhau
ta có:
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2-y^2+2z^2}{4-9+32}=\frac{108}{27}=4\)
\(\Rightarrow\)\(\frac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x=4\)
\(\frac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow y=6\)
\(\frac{z^2}{16}=4\Rightarrow z^2=64\Rightarrow z=8\)
vậy....
k mik nhé
hok tốt
Bài làm
Vì \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
=> \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{64}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
Ta có: \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{64}\Rightarrow\frac{x^2-y^2+2z^2}{4-9+64}=\frac{108}{59}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{108}{59}\Rightarrow x=\frac{216}{59}\\\frac{y}{3}=\frac{108}{59}\Rightarrow y=\frac{324}{59}\\\frac{z}{4}=\frac{108}{59}\Rightarrow z=\frac{432}{59}\end{cases}}\)
Vậy \(x=\frac{216}{59}\)
\(y=\frac{324}{59}\)
\(z=\frac{432}{59}\)
ĐẶT
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\)
\(\Rightarrow x=2k,y=3k,z=4k\)
khi đó\(x^2-y^2+2z^2=108\)trở thành:
\(4k^2-9k^2+64k^2=108\)
\(\Rightarrow k^2\left(4-9+64\right)=108\)
\(\Rightarrow k^2\cdot59=108\)
\(\Rightarrow k^2=2\)
\(\Rightarrow k=\sqrt{2}\)
\(\Rightarrow\hept{\begin{cases}a=\sqrt{2}\cdot2\\b=\sqrt{2}\cdot3\\c=\sqrt{2}\cdot4\end{cases}}\)
Ta có : \(\left(\frac{x}{2}\right)^2=\left(\frac{y}{3}\right)^2=\left(\frac{z}{4}\right)^2\Rightarrow\frac{x^2}{4}=\frac{y}{9}^2=\frac{2z^2}{32}\)
Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2-y^2+2z^2}{4-9+32}=\frac{108}{27}=4\)
\(\Rightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
\(y^2=36\Rightarrow\orbr{\begin{cases}y=6\\y=-6\end{cases}}\)
\(z^2=32:2\cdot4=64\Rightarrow\orbr{\begin{cases}z=8\\z=-8\end{cases}}\)
Vậy ta tìm được 2 cặp:\(\hept{\begin{cases}x=8;y=6;z=8\\x=-4;y=-6;z=-8\end{cases}}\)
