phân số 2015/2014 phải ko, bn k cho mk nhé , mk trả lời câu hỏi của bạn rồi đấy

Ta có:

\(M=\frac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{xyz-x^3+xyz-y^3+xyz-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{3xyz-x^3-y^3-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)

\(-M=\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)

Xét đẳng thức phụ:

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

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

\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-abc-ac\right)\)

\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

Thay vào -M ta có:

\(-M=\frac{\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{1}{2}\left(x+y+z\right)\Rightarrow M=-\frac{1}{2}\left(x+y+z\right)\)

Giờ thay: \(x=2014^{2015}-20142015;y=20142015-2015^{2014};z=2015^{2014}-2014^{2015}\)

Ta có:

\(M=-\frac{1}{2}\left(2014^{2015}-20142015+20142015-2015^{2014}+2015^{2014}-2014^{2015}\right)=0\)

mk nè,k đi

Ai giải hộ mik bài này đi mình K cho

Ta có:

2014/2015=20142014/20152015

Vì 20142014<20142015 nên 20142014/20152015<20142015/20152015

=> Nó <

20142014/20152015=2014.10001/2015.10001

=2014/2015

Ta có:

2015²⁰¹⁷ + 2017²⁰¹⁵

= 2015²⁰¹⁷ + 1 + 2017²⁰¹⁵ - 1

= (2015²⁰¹⁷ + 1²⁰¹⁷) + (2017²⁰¹⁵ - 1²⁰¹⁵)

Do 2015²⁰¹⁷ + 1²⁰¹⁷ luôn chia hết cho 2015 + 1 = 2016; 2017²⁰¹⁵ - 1²⁰¹⁵ luôn chia hết cho 2017 - 1 = 2016

=> (2015²⁰¹⁷ + 1²⁰¹⁷) + (2017²⁰¹⁵ - 1²⁰¹⁵) chia hết cho 2016

=> 2015²⁰¹⁷ + 2017²⁰¹⁵ chia hết cho 2016 (đpcm)

TAU KHONG BIET