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if a<b,bcz of a^b=b^c so b>c c<d d>e e<f f>g g<a bcz of g<a and a<b so g<b (not possible)
Same with a>b ,so a=b.
Do again multiple time ,we get a=b=c=d=e=f so bcs f^g=g^a,so f^g=g^f so g=f.
So totally ,we get a=b=c=d=e=f=g.
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Áp dụng dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{g}=\frac{a+b+c+d+e}{b+c+d+e+g}\)
=> \(\left(\frac{a}{b}\right)^{404}.\left(\frac{b}{c}\right)^{404}.\left(\frac{c}{d}\right)^{404}.\left(\frac{d}{e}\right)^{404}.\left(\frac{e}{g}\right)^{404}\)
\(=\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}.\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}.\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}.\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}.\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}\)
=> \(\left(\frac{abcde}{bcdeg}\right)^{404}=\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404+404+404+404}\)
=> \(\frac{a^{404}}{g^{404}}=\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{2020}\)
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tớ biết tớ ....................................................................chết liền!
Vì \(\frac{a}{b}>\frac{c}{d}>\frac{e}{g}\Rightarrow\hept{\begin{cases}ad>bc\\cg>ed\end{cases}}\Rightarrow\hept{\begin{cases}ad-bc>0\\cg-ed>0\end{cases}}\Rightarrow\hept{\begin{cases}ad-bc\ge1\\cg-ed\ge1\end{cases}}\)(Do a, b, c, d, g nguyên dương)
Ta có: \(d=d\left(ag-be\right)=adg-bed=\left(adg-bcg\right)+\left(bcg-bed\right)\)
\(=g\left(ad-bc\right)+b\left(cg-ed\right)\ge g.1+b.1=b+g\)
Ta có điều phải chứng minh