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

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)

\(\Leftrightarrow\left(a^2+2ab+b^2\right)c+ab\left(a+b\right)+c^2\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0

=> Hoặc a=-b hoặc b=-c hoặc c=-a

Ko mất tổng quát, g/s a=-b

a) Ta có: vì a=-b thay vào ta được:

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)

\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)

=> đpcm

b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)

=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)

Đặt \(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=3k\\b+c=4k\\c+a=5k\end{matrix}\right.\Rightarrow2\cdot\left(a+b+c\right)=12k\Rightarrow a+b+c=6k\)

\(\Rightarrow\left\{{}\begin{matrix}c=3k\left(1\right)\\a=2k\left(2\right)\\b=k\left(3\right)\end{matrix}\right.\)

Thay \(\left(1\right),\left(2\right),\left(3\right)\) vào BT ta có:

\(M=10\cdot2k+k-7\cdot3k+2017\)

\(M=20k+k-21k+2017\)

\(M=21k-21k+2017\)

\(M=2017\)

Vậy \(M=2017\)

Đặt \(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=3k\\b+c=4k\\c+a=5k\end{matrix}\right.\)

\(\Rightarrow2\left(a+b+c\right)=12k\)

\(\Rightarrow a+b+c=6k\) \(\Rightarrow\left\{{}\begin{matrix}a=2k\\b=k\\c=3k\end{matrix}\right.\)

Thay a = 2k , b = k , c= = 3k vào biểu thức M , ta có :

M = 10.2k + k - 7.3k + 2017 = 21k - 21k + 2017 = 2017

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=\dfrac{a+b+b+c+c+a}{3+4+5}=\dfrac{a+b+b+c}{3+4}=\dfrac{b+c+c+a}{4+5}=\dfrac{a+b+c+a}{3+5}=\dfrac{a+b+c}{6}=\dfrac{a+2b+c}{7}=\dfrac{b+2c+a}{9}=\dfrac{2a+b+c}{8}\)

Tiếp tục áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a+b+c}{6}=\dfrac{a+2b+c}{7}=\dfrac{b+2c+a}{9}=\dfrac{2a+b+c}{8}=\dfrac{2a+b+c-a-b-c}{8-6}=\dfrac{a}{2}\left(1\right)\)

\(\dfrac{a+b+c}{6}=\dfrac{a+2b+c}{7}=\dfrac{b+2c+a}{9}=\dfrac{2a+b+c}{8}=\dfrac{a+2b+c-a-b-c}{7-6}=\dfrac{b}{1}\left(2\right)\)

\(\dfrac{a+b+c}{6}=\dfrac{a+2b+c}{7}=\dfrac{b+2c+a}{9}=\dfrac{2a+b+c}{8}=\dfrac{b+2c+a-a-b-c}{9-6}=\dfrac{c}{3}\left(3\right)\)

Từ (1);(2) và (3) ta có: \(\dfrac{a}{2}=\dfrac{b}{1}=\dfrac{c}{3}\)

Đặt: \(\dfrac{a}{2}=\dfrac{b}{1}=\dfrac{c}{3}=t\Leftrightarrow\left\{{}\begin{matrix}a=2t\\b=t\\c=3t\end{matrix}\right.\)

Thay vào \(M\) ta có: \(M=10a+b-7c+2017=20t+t-21t+2017=2017\)

oh no bài thứ nhất là dạng chứng minh cs đúng ko ,

ko thể nào là dạng tìm a,b,c đc-.-

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2022}\)

hay \(\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc=abc\)

\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+2abc=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

-Xét a + b = 0 => P = 2022^2021

Bạn xét tương tự với b + c = 0 và c + a = 0 dc P = 2022^2021 nhé

a+bab+a+bc(a+b+c)=0a+bab+a+bc(a+b+c)=0

(a+b)[ab+bc+ca+c2abc(a+b+c)]=0(a+b)[ab+bc+ca+c2abc(a+b+c)]=0

 ⇔ a=−b

 ⇔ b=−c

 ⇔ c=−a

Thay vào P từng cái rồi tính tiếp nhé

\(\dfrac{a}{2021-c}+\dfrac{b}{2021-a}+\dfrac{c}{2021-b}\\ =\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ =\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

Vì \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\Rightarrow A.ko.phải.số.nguyên\)