\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c};c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{c^3+b^3+d^3}\left(1\right)\\ \text{Đặt }\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;b=ck;c=dk\\ \Rightarrow a=bk=ck^2=dk^3\\ \Rightarrow\dfrac{a}{d}=k^3\\ \text{Mà }\dfrac{a}{b}=k\Rightarrow\dfrac{a^3}{b^3}=k^3\\ \Rightarrow\dfrac{a}{d}=\dfrac{a^3}{b^3}\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)

Ta có: \(ac=b^2\)

=>\(\frac{a}{b}=\frac{b}{c}\)

Ta có: \(ab=c^2\)

=>\(\frac{b}{c}=\frac{c}{a}\)

=>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k\)

=>\(\begin{cases}c=ak\\ b=ck=ak\cdot k=ak^2\\ a=bk=ak^2\cdot k=ak^3\end{cases}\Rightarrow\begin{cases}c=ak\\ b=ak^2\\ a\left(k^3-1\right)=0\end{cases}\)

=>\(\begin{cases}c=ak\\ b=ak^2\\ k^3-1=0\end{cases}\Rightarrow\begin{cases}k=1\\ c=a\cdot1=a\\ b=a\cdot1^2=a\end{cases}\)

=>a=b=c

\(P=\frac{a^{555}}{b^{222}\cdot c^{333}}+\frac{b^{555}}{c^{222}\cdot a^{333}}+\frac{c^{555}}{a^{222}\cdot b^{333}}\)

\(=\frac{a^{555}}{a^{222}\cdot a^{333}}+\frac{a^{555}}{a^{222}\cdot a^{333}}+\frac{a^{555}}{a^{222}\cdot a^{333}}\)

=1+1+1

=3

Lời giải:

Ta thấy:
$(ab+cd)(ac+bd)=ad(c^2+b^2)+bc(a^2+d^2)$

$=(ad+bc)t$

Mà: 

$2(t-ab-cd)=(a-b)^2+(c-d)^2>0$ nên $t> ab+cd$

Tương tự: $t> ac+bd$

Kết hợp $(ab+cd)(ac+bd)=(ad+bc)t$ nên:

$ab+cd> ad+bc, ac+bd> ad+bc$

Nếu $ab+cd, ac+bd$ đều thuộc $P$. Do $ad+bc$ là ước của $ab+cd$ hoặc $ac+bd$. Điều này vô lý 

Do đó ta có đpcm.

  là số chẵn

Cc

b^2=ac

=>b/a=c/b=k

=>b=ak; c=bk=ak*k=ak^2

\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+a^2k^2}{a^2k^2+a^2k^4}=\dfrac{1}{k^2}\)

\(\dfrac{a}{c}=\dfrac{a}{ak^2}=\dfrac{1}{k^2}\)

=>\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)