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Giải:
Gọi 3 số đó là a, b, c
Ta có: \(\frac{a}{\frac{2}{5}}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{1}{6}}\) và \(a^2+b^2+c^2=24309\)
Đặt \(\frac{a}{\frac{2}{5}}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{1}{6}}=k\Rightarrow\left\{\begin{matrix}a=\frac{2}{5}k\\b=\frac{3}{4}k\\c=\frac{1}{6}k\end{matrix}\right.\)
Mà \(a^2+b^2+c^2=24309\)
\(\Rightarrow\frac{4}{25}k^2+\frac{9}{16}k^2+\frac{1}{36}k^2=24309\)
\(\Rightarrow\frac{2701}{3600}k^2=24309\)
\(\Rightarrow k^2=32400\)
\(\Rightarrow\left\{\begin{matrix}k=180\\k=-180\end{matrix}\right.\)
+) \(k=180\Rightarrow a=72;b=135;c=30\)
\(\Rightarrow a+b+c=237\)
+) \(k=-180\Rightarrow a=-72;b=-135;c=-30\)
\(\Rightarrow a+b+c=-237\)
Vậy \(\left[\begin{matrix}a+b+c=237\\a+b+c=-237\end{matrix}\right.\)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0.\)
\(1+\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}-4+\frac{x+349}{5}=0\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
Study well
\(\Rightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Mà \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
Nên \(x+329=0\Rightarrow x=-329\)
Vậy \(x=-329\)
Chúc bạn học tốt !!!
