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: ...

a, \(a^3+b^3+c^3=3abc\)

⇔\(a^3+b^3+c^3-3abc=0\)

⇔\(\left(a+b\right)^3+c^3-3abc-3a^2b-3ab^2=0\)

⇔\(\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)=0\)

⇔\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc-3ab\right)=0\)

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

⇒\(a^2+b^2+c^2-ab-bc-ac=0\left(a+b+c\ne0\right)\)

⇔\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

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

⇔\(a=b=c\)

⇒\(=\frac{a^{2016}+a^{2016}+a^{2016}}{\left(a+a+a\right)^{2016}}=\frac{3a^{2016}}{3^{2016}\cdot a^{2016}}=\frac{1}{3^{2015}}\)

b/ \(n^2+4n+2013=k^2\) (\(k\in N\))

\(\Leftrightarrow\left(n+2\right)^2+2009=k^2\)

\(\Leftrightarrow k^2-\left(n+2\right)^2=2009\)

\(\Leftrightarrow\left(k-n-2\right)\left(k+n+2\right)=2009=1.2009=7.287=41.49\)

Do \(k-n-2< k+n+2\) nên ta chỉ cần xét 3 trường hợp:

\(\left\{{}\begin{matrix}k-n-2=1\\k+n+2=2009\end{matrix}\right.\) \(\Rightarrow2n+4=2008\Rightarrow n=1002\)

\(\left\{{}\begin{matrix}k-n-2=7\\k+n+2=287\end{matrix}\right.\) \(\Rightarrow n=138\)

\(\left\{{}\begin{matrix}k-n-2=41\\k+n+2=49\end{matrix}\right.\) \(\Rightarrow n=2\)

Vậy \(n=\left\{2;138;1002\right\}\)

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

Ta có : \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

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

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-cb\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-cb-3ab\right)=0\)

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

Vi a,b,c khác 0 Nên : ​\(a^2+b^2+c^2-ab-bc-ac=0\)

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

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

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

\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)

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

<=> a = b = c

Vậy \(N=\frac{a^{2016}+b^{2016}+c^{2016}}{\left(a+b+c\right)^{2016}}=\frac{a^{2016}+a^{2016}+a^{2016}}{\left(a+a+a\right)^{2016}}=\frac{3.a^{2016}}{3^{2016}.a^{2016}}=\frac{1}{3^{2015}}\)

cảm ơn so tài kết thúc

 ai win vậy  

a) để 5/n-1 là số nguyên thì 5 chia hết cho n-1

=> n-1 thuộc Ư(5)=( 1, -1, 5, -5)

ta có

n-1=1=>n=2

n-1=-1=>n=0

n-1=5=>n=6

n-1=-5=>n=-4

mà n là số tự nhiên => n thuộc 2,0,6

máy mik bị lỗi bàn phím nên phải gõ ngoặc khác thay thế TvT, sorry nghen

b) M=(1-1000/2016) *...*(1-2016/2016)*(1-2017/2016)

=>M=(1-1000/2016)*.....*0*(1-2017/2016)

=>M=0

bài này dễ vào TH 0,5 điểm trong bài thi

nghe có vẻ khó nhưng chú ý 1 chút là có thể làm được

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^{2016}}{c^{2016}}=\frac{b^{2016}}{d^{2016}}\)\(\Rightarrow\left(\frac{a^{2016}}{c^{2016}}\right)^{2017}=\left(\frac{b^{2016}}{d^{2016}}\right)^{2017}\)

áp dụng t/c dãy t/s = nhau

\(\Rightarrow\left(\frac{a^{2016}}{c^{2016}}\right)^{2017}=\left(\frac{b^{2016}}{d^{2016}}\right)^{2017}=\)\(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)

biến đổi tiếp cái kia tương tự rồi suy ra chúng = nhau nhé

1,

Ta có: \(x^2\ge0;\left|y-13\right|\ge0\)

\(\Rightarrow x^2+\left|y-13\right|\ge0\)

\(\Rightarrow x^2+\left|y-13\right|+14\ge14\)

\(\Rightarrow\frac{1}{x^2+\left|y-13\right|+14}\le\frac{1}{14}\)

\(\Rightarrow P=\frac{12}{x^2+\left|y-13\right|+14}\le\frac{12}{14}=\frac{6}{7}\)

Dấu "=" xảy ra khi x = 0, y = 13

Vậy P_min = 6/7 khi x = 0, y = 13

2, \(P=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=1+\frac{7}{n-5}\)

Để P có GTLN thì\(\frac{7}{n-5}\) có GTLN => n - 5 có GTNN và n - 5 > 0 => n = 6

3,

Ta có: \(10\le n\le99\)

\(\Rightarrow20\le2n\le198\)

\(\Rightarrow2n\in\left\{36;64;100;144;196\right\}\)

\(\Rightarrow n\in\left\{18;32;50;72;98\right\}\)

\(\Rightarrow n+4\in\left\{22;36;50;72;98\right\}\)

Ta thấy chỉ có 36 là số chính phương 

Vậy n = 32

4,

ÁP dụng TCDTSBN ta có:

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c+b+c-a+a+c-b}{c+a+b}=\frac{a+b+c}{a+b+c}=1\) (vì a+b+c khác 0)

\(\Rightarrow\hept{\begin{cases}\frac{a+b-c}{c}=1\\\frac{b+c-a}{a}=1\\\frac{a+c-b}{b}=1\end{cases}\Rightarrow\hept{\begin{cases}a+b-c=c\\b+c-a=a\\a+c-b=b\end{cases}\Rightarrow}\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}}\)

\(\Rightarrow B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}\cdot\frac{a+c}{c}\cdot\frac{b+c}{b}=\frac{2c}{a}\cdot\frac{2b}{c}\cdot\frac{2a}{b}=\frac{8abc}{abc}=8\)

Vậy B = 8 

5,

Ta so sánh 32²⁷ và 18³⁹

32²⁷ =(2⁵)²⁷ = 2¹³⁵ < 2¹⁵⁶ = (2⁴)³⁹ = 16³⁹ < 18³⁹

Vậy (-32)²⁷ > (-18)³⁹

6, làm tương tự 2