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Khác số chút thoyy.

\(P=\dfrac{1}{bc\left(b+c\right)+2023}+\dfrac{1}{ca\left(c+a\right)+2023}+\dfrac{1}{ab\left(a+b\right)+2023}\left(abc=2023\right)\)

\(\Leftrightarrow P=\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{1}{ca\left(c+a\right)+abc}+\dfrac{1}{ab\left(a+b\right)+abc}\)

\(\Leftrightarrow P=\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ca\left(a+b+c\right)}+\dfrac{1}{ab\left(a+b+c\right)}\)

\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\)

\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{a^2bc+b^2ca+c^2ab}{\left(abc\right)^2}\right]\)

\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{abc\left(a+b+c\right)}{\left(abc\right)^2}\right]\)

\(\Leftrightarrow P=\dfrac{1}{abc}=\dfrac{1}{2023}\)

Ta có:\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

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

Ta có:\(\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\frac{a\cdot a^2+a\cdot a^2+a\cdot a^2}{a^3+a^3+a^3}\)\(\Rightarrow\frac{3a^3}{3a^3}=1\)

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

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

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

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

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

<=> a = b = c

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

có gì khó thì hỏi mình qua số điện thoại hỗ trợ

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\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)

\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)

Lý do gì mà người ra đề lại chọn 1 con số xấu phi lý như 9 ở đây nhỉ? Vì con số này là ko có ý nghĩa (2, 3, 4, 6 hay 9 gì thì cách giải đều giống nhau, nhưng việc chọn 9 khiến kết quả xấu khủng khiếp)

\(9=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le3\sqrt{2}\)

\(P=\dfrac{ab}{a+b+3}\le\dfrac{\left(a+b\right)^2}{4\left(a+b+3\right)}\)

Đặt \(a+b=x\Rightarrow0< x\le3\sqrt{2}\)

\(4P\le\dfrac{x^2}{x+3}=\dfrac{x^2}{x+3}+6-6\sqrt{2}-6+6\sqrt{2}\)

\(4P\le\dfrac{x^2+\left(6-6\sqrt{2}\right)x+18-18\sqrt{2}}{x+3}-6+6\sqrt{2}\)

\(4P\le\dfrac{\left(x-3\sqrt{2}\right)\left(x+6-3\sqrt{2}\right)}{x+3}-6+6\sqrt{2}\le-6+6\sqrt{2}\)

\(P\le\dfrac{-3+3\sqrt{2}}{2}\)

\(P_{max}=\dfrac{-3+3\sqrt{2}}{2}\) khi \(x=3\sqrt{2}\) hay \(a=b=\dfrac{3\sqrt{2}}{2}\)

\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{a+c}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ac}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)

\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\Rightarrow M=1\)

Lời giải:

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

\(\Leftrightarrow c(a+b)=a(b+c)=b(c+a)\)

\(\Leftrightarrow ac+bc=ab+ac=bc+ab\Leftrightarrow ab=bc=ac\)

\(\Rightarrow a=b=c\) (do $a,b,c>0$)

$\Rightarrow M=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1$