Ta có 

a²+b²+c² = ab+bc+ca

<=> 2(a²+b²+c²)= 2(ab+bc+ca)

<=> (a - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²) = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

<=> a = b = c

Thế vào pt thứ (2) ta được

a⁸ + b⁸ + c⁸ = 3

<=> 3a⁸ = 3

<=> a⁸ = 1

<=> a = b = c = 1(3) hoặc a = b = c = - 1(4)

Từ (3) => P = 1 + 1 - 1 = 1

Từ (4) => P = - 1 + 1 + 1 = 1

\(a^2+b^2+c^2=ab+bc+ca\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

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

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

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\)\(\Rightarrow a-b=b-c=c-a=0\)

\(\Rightarrow P=\left(a-b\right)^{2015}+\left(b-c\right)^{2016}+\left(c-a\right)^{2017}=0\)

cảm ơn bạn nha

a² + b² + c² = ab + bc + ca 

=>2.(a²+b²+c²)=2.(ab+bc+ca)

<=>a²+2b²+2c²=2ab+2bc+2ca

<=>2a²+2b²+2c²-2ab-2bc-2ca=0

<=>a²-2ab+b²+b²-2bc+c²+c²-2ca+a²=0

<=>(a-b)²+(b-c)²+(c-a)²=0

<=>a-b=0 và b-c=0 và c-a=0

<=>a=b và b=c và c=a

=> a=b=c

mà a;b;c khác 0 nên

P=1+1+1=3

a² + b² + c² = ab + bc + ca => 2. (a² + b² + c² )= 2.( ab + bc + ca) 

<=> (a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²) = 0 

<=> (a - b)² + (b - c)² + (c - a)² = 0  <=> (a - b)²  =  (b - c)²  =  (c - a)² = 0  (Vì (a - b)² \(\ge\) 0;  ( b - c)² \(\ge\)0 ;  (c - a)² \(\ge\) 0

<=> a = b = c

=> \(P=\frac{a^4}{a^4}+\frac{b^4}{b^4}+\frac{a^{2016}}{a^{2016}}=1+1+1=3\)

a) \(a^2+b^2+c^2=ab+bc+ac\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

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

Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(c-a\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)

\(\Leftrightarrow a=b=c\)

a. \(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ab-2ca=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)

c) \(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}+\frac{x+2}{2018}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}+\frac{x+2018}{2}\)

Ta có VT + 4 = VP + 4

VT + 4 = \(\left(\frac{x+5}{2015}+1\right)+\left(\frac{x+4}{2016}+1\right)+\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+2}{2018}+1\right)\) 

\(=\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}+\frac{x+2020}{2018}\)

\(=\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)\)

VP + 4 = \(\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)+\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}\right)\)

\(=\frac{x+2020}{5}+\frac{x+2020}{4}+\frac{x+2020}{3}+\frac{x+2020}{2}\)

\(=\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)

Khi đó \(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)

=> \(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

Vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)

=> x + 2020 = 0

=> x = -2020

Em tham khảo cách làm tại link: Câu hỏi của Cao Chi Hieu - Toán lớp 9 - Học toán với OnlineMath

a²+b²+c²=ab+bc+ca

<=>2a²+2b²+2c²=2ab+2bc+2ca

<=>(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)=0

<=>(a-b)²+(b-c)²+(c-a)²=0

<=>a=b=c

mà a+b+c=3<=>a=b=c=1

=>P=0

P=2017 chứ bạn