\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\\\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{a}\end{cases}}\)

\(\Leftrightarrow a=b=c\)

Vậy P =1

Ta có: \(\frac{ab + ac}{2} = \frac{bc + ba}{3} = \frac{ca + cb}{4}\)

=>\(\frac{a(b + c)}{2} = \frac{b(c + a)}{3} = \frac{c(a + b)}{4}\)

=>\(\frac{a(b + c)}{2abc} = \frac{b(c + a)}{3abc} = \frac{c(a + b)}{4abc}\)

=>\(\frac{b + c}{2bc} = \frac{c + a}{3ca} = \frac{a + b}{4ab}\)

=>\(\frac{b}{2bc} + \frac{c}{2bc} = \frac{c}{3ca} + \frac{a}{3ca} = \frac{a}{4ab} + \frac{b}{4ab}\)

=>\(\frac{1}{2c}+\frac{1}{2b}=\frac{1}{3a}+\frac{1}{3c}=\frac{1}{4b}+\frac{1}{4a}\)

=>\(12 \cdot \left(\frac{1}{2c} + \frac{1}{2b}\right) = 12 \cdot \left(\frac{1}{3a} + \frac{1}{3c}\right) = 12 \cdot \left(\frac{1}{4b} + \frac{1}{4a}\right)\)

=>\(\frac{6}{c}+\frac{6}{b}=\frac{4}{a}+\frac{4}{c}=\frac{3}{b}+\frac{3}{a}\quad(*)\)

\(\frac{6}{c} + \frac{6}{b} = \frac{4}{a} + \frac{4}{c}\)

=>\(\frac{6}{c}-\frac{4}{c}+\frac{6}{b}=\frac{4}{a}\)

=>\(\frac{2}{c}+\frac{6}{b}=\frac{4}{a}\quad(1)\)

Ta có: \(\frac{4}{a} + \frac{4}{c} = \frac{3}{b} + \frac{3}{a}\)

=>\(\frac{4}{a}-\frac{3}{a}+\frac{4}{c}=\frac{3}{b}\)

=>\(\frac{1}{a}+\frac{4}{c}=\frac{3}{b}\implies\frac{3}{b}-\frac{1}{a}=\frac{4}{c}\quad(2)\)

\(\frac{6}{c} + \frac{6}{b} = \frac{3}{b} + \frac{3}{a}\)

=>\(\frac{6}{b}-\frac{3}{b}+\frac{6}{c}=\frac{3}{a}\)

=>\(\frac{3}{b}+\frac{6}{c}=\frac{3}{a}\implies\frac{1}{b}+\frac{2}{c}=\frac{1}{a}\quad(3)\)

Từ (1),(3) suy ra \(\frac{2}{c} + \frac{6}{b} = 4 \left(\frac{1}{b} + \frac{2}{c}\right)\)

=>\(\frac{2}{c}+\frac{6}{b}=\frac{4}{b}+\frac{8}{c}\)

=>\(\frac{-6}{c}=\frac{-2}{b}\)

=>\(\frac{6}{c}=\frac{2}{b}\)

=>\(\frac{c}{6}=\frac{b}{2}\)

=>\(\frac{c}{15}=\frac{b}{5}\) (4)

=>\(\frac{1}{b}=\frac{3}{c}\)

=>\(\frac{1}{a} = \frac{3}{c} + \frac{2}{c} = \frac{5}{c}\)

=>\(\frac{a}{1}=\frac{c}{5}\)

=>\(\frac{a}{3}=\frac{c}{15}\) (5)

Từ (4),(5) suy ra \(\frac{a}{3}=\frac{b}{5}=\frac{c}{15}\) (ĐPCM)

\(\frac{ab+ac}{2}=\frac{bc+ab}{3}=\frac{ca+bc}{4}\)

( ta lần lược lấy - (1) + (2) + (3) = (1) - (2) + (3) = (1) + (2) - (3) được)

\(=\frac{2bc}{5}=\frac{2ca}{3}=\frac{2ab}{1}\)

Ta thấy rằng a,b,c không thể = 0 vì như vậy thì a + b + c \(\ne69\)

\(\Rightarrow\hept{\begin{cases}a=\frac{c}{5}\\b=\frac{c}{3}\end{cases}}\)

Thế vào: a + b + c = 69

\(\Leftrightarrow\frac{c}{5}+\frac{c}{3}+c=69\)

\(\Rightarrow c=45\)   

\(\Rightarrow\hept{\begin{cases}a=9\\b=15\end{cases}}\)  

Dùng tính chất dãy tỉ số bằng nhau mà làm

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

Vì $abc\ne0$ nên từ $a+b+c=\dfrac1a+\dfrac1b+\dfrac1c$

suy ra $abc(a+b+c)=ab+bc+ca$.

Xét hiệu hai vế cần chứng minh:

$b(a^2-bc)(1-ac)-a(1-bc)(b^2-ac)$

$=-(a-b)\left[abc(a+b+c)-(ab+bc+ca)\right]$

$=-(a-b)\cdot0$

$=0$

Suy ra $\boxed{b(a^2-bc)(1-ac)=a(1-bc)(b^2-ac)}$

Ta có $(a+b+c)^2=a^2+b^2+c^2$

$\Rightarrow 2(ab+bc+ca)=0$

$\Rightarrow ab+bc+ca=0$

Ta có $\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}
=\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}$

Mà $a^3b^3+b^3c^3+c^3a^3-3a^2b^2c^2$

$=(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2-abc(a+b+c))$

$=0$

Do đó $a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2$

Suy ra $\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}$

$=\dfrac{3a^2b^2c^2}{a^3b^3c^3}$

$=\boxed{\dfrac3{abc}}$

\(\frac{a.b}{a+b}=\frac{b.c}{b+c}=\frac{c.a}{c+a}\)

\(\Rightarrow\frac{a+b}{a.b}=\frac{b+c}{b.c}=\frac{c+a}{c.a}\) (vì a;b;c khác 0)

\(=\frac{a}{a.b}+\frac{b}{a.b}=\frac{b}{b.c}+\frac{c}{b.c}=\frac{c}{c.a}+\frac{a}{c.a}\)

\(=\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)

=> a = b = c

\(P=\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\frac{a.a^2+a.a^2+a.a^2}{a^3+a^3+a^3}=\frac{a^3+a^3+a^3}{a^3+a^3+a^3}=1\)

ab² hay là a²b²

Là a.b^2 nhé