`Answer:`

Có `a^2.(b+c)=b^2.(a+c)`

`<=>a^2.b+a^2.c-ab^2-b^2.c=0`

`<=>ab.(a-b)+c.(a^2-b^2)=0`

`<=>(a-b)(ab+c(a+b))=0`

`<=>(a-b)(ab+ac+bc)=0`

`<=>ab+ac+bc=0`

Lúc này  `P=c^2.(a+b)=c.(ac+bc)=c.(-ab)=-abc`

Mà `a^2.(b+c)=a.(ab+ac)=a.(-bc)=-abc=2022`

Vậy `P=2022`

Ta có :\(a^2+b=b^2+c\Rightarrow\left(a-b\right)\left(a+b\right)=c-b\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-\left(a-b\right)=c-b-a+b\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=c-a\)

Tương tự \(\hept{\begin{cases}\left(b-c\right)\left(b+c-1\right)=a-b\\\left(a-c\right)\left(a+c-1\right)=c-b\end{cases}}\)

Nhận vế với vế của các đẳng thức trên ta được :

\(\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b-1\right)\left(b+c-1\right)\left(a+c-1\right)=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(\Rightarrow\left(a+b-1\right)\left(b+c-1\right)\left(a+c-1\right)=1\)

Nhân khai triển tử và mẫu của B, thấy ab + bc + ca thì thay bằng 1

Ta có:

\(\left\{{}\begin{matrix}a^2+b=b^2+c\\b^2+c=c^2+a\\a^2+b=c^2+a\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a^2-b^2=c-b\\b^2-c^2=a-c\\a^2-c^2=a-b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=c-b\\\left(b-c\right)\left(b+c\right)=a-c\\\left(a-c\right)\left(a+c\right)=a-b\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{c-b}{a-b}\\b+c=\dfrac{a-c}{b-c}\\a+c=\dfrac{a-b}{a-c}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b-1=\dfrac{c-a}{a-b}\\b+c-1=\dfrac{a-b}{b-c}\\a+c-1=\dfrac{c-b}{a-c}\end{matrix}\right.\)

\(\Rightarrow T=\left(a+b-1\right)\left(b+c-1\right)\left(a+c-1\right)\)

\(=\dfrac{\left(c-a\right)\left(a-b\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)

Tham khảo:

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