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![](https://rs.olm.vn/images/avt/0.png?1311)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{g}=\frac{a+b+c+d+e}{b+c+d+e+g}\)
=> \(\left(\frac{a}{b}\right)^{404}.\left(\frac{b}{c}\right)^{404}.\left(\frac{c}{d}\right)^{404}.\left(\frac{d}{e}\right)^{404}.\left(\frac{e}{g}\right)^{404}\)
\(=\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}.\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}.\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}.\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}.\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404}\)
=> \(\left(\frac{abcde}{bcdeg}\right)^{404}=\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{404+404+404+404}\)
=> \(\frac{a^{404}}{g^{404}}=\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^{2020}\)
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\) (1)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\) (2)
\(d^2=ce\Rightarrow\frac{c}{d}=\frac{d}{e}\) (3)
\(e^2=dg\Rightarrow\frac{d}{e}=\frac{e}{g}\) (4)
Từ (1),(2),(3),(4) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{g}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{g}=\frac{a+b+c+d+e}{b+c+d+e+g}\)
Ta có: \(\frac{a}{b}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (5)
\(\frac{b}{c}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (6)
\(\frac{c}{d}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (7)
\(\frac{d}{e}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (8)
\(\frac{e}{g}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (9)
Nhân (5),(6),(7),(8),(9) vế với vế:
\(\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}\cdot\frac{d}{e}\cdot\frac{e}{g}=\frac{a}{g}=\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^5\) (đpcm)
P/s: Mk nghĩ đề là c/m: a/g = (a+b+c+d+e/b+c+d+e+g)^5
đề đúng k vậy