A=\(2016^2=2016.2016\)

B=\(2015.2017=(2015+1)(2017-1)=2016.2016\)

=> A=B = 2016.2016

\(B=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1< 2016^2=A\)

Có:

\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

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

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)

Thay \(a=b=c\) vào \(A\), ta được:

\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)

\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)

\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)

\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)

\(=\dfrac{3}{2017^2}\)

Vậy: ...

Giúp tôi với !!

Cô ơi em có cách khác ạ :)

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)

Dấu "=" xảy ra tại x=y=z=0

Khi đó T=0

Ta có: 

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

<=> \(\left(a^2+b^2+c^2\right)\)\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\left(a^2+b^2+c^2\right)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

<=> \(x^2+y^2+z^2=\left(a^2+b^2+c^2\right)\frac{x^2}{a^2}+\left(a^2+b^2+c^2\right)\frac{y^2}{b^2}+\left(a^2+b^2+c^2\right)\frac{z^2}{c^2}\)

<=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)

vì a, b , c khác 0 nên \(\frac{\left(b^2+c^2\right)}{a^2};\frac{\left(c^2+a^2\right)}{b^2};\frac{\left(b^2+a^2\right)}{c^2}\ne0\)

\(\frac{\left(b^2+c^2\right)}{a^2}x^2\ge0;\frac{\left(a^2+c^2\right)}{b^2}y^2\ge0;\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x, y, z

=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x; y; z

Do đó: \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)

=> x = y = z = 0

Vậy T = 0 

Ta có: a³ + b³ + c³ = 3abc 

  \(\Leftrightarrow\)a³ + b³ + c³ - 3abc = 0

  \(\Leftrightarrow\)(a + b)³ + c³ - 3ab² - 3a²b - 3abc = 0

  \(\Leftrightarrow\)(a + b + c)[(a + b)² - c(a + b) + c² ] - 3ab(a + b + c) = 0

  \(\Leftrightarrow\)(a + b + c)(a² + 2ab + b² - ac - bc + c² - 3ab) = 0

  \(\Leftrightarrow\)(a + b + c)(a² + b² + c² - ab - bc - ca) = 0

Vì a + b + c khác 0 nên

    a² + b² + c² - ab - bc - ca = 0

\(\Leftrightarrow\)2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

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

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

 N = \(\frac{a^{2016}+b^{2016}+c^{2016}}{\left(a+b+c\right)^{2016}}\)= 1

\(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1\)

\(< 2016^2=B\)

Nên A<B

\(B=2016^2\)

\(\Rightarrow B=\left(2017-1\right)^2\)

\(\Rightarrow B=2017^2-4034+1=2017^2-4033\)(1)

Lại Có :

\(A=2015.2017=\left(2017-2\right).2017\)

\(\Rightarrow A=2017^2-4034\)(2)

Từ (1) và (2) =>  B>A

Cho a+b+c=0 và a^2+b^2+c^2=14.Tính P=a^4+b^4+c^4

(a+b+c)^2=0=>a^2+b^2+c^2+2(ab+bc+ac)=0=>2(ab+bc+ac)=-14=>(ab+ac+bc)^2=49                                                                        phân tích (ab+ac+cb)^2 ta được (ab)^2+(bc)^2+(ac)^2=49                                                                                                              đặt N=  a^2+b^2+c^2=14=>   N^2=196                                                                                                                                           phân tích N^2 rồi thế (ab)^2+(bc)^2+(ac)^2=49 vào N^2 sẽ có kết quả của a^4+b^4+c^4