a/ Với \(x=2016\Rightarrow2017=x+1\)

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+2025\)

\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2025\)

\(A=2025-x=9\)

b/ Với \(x=-1\Rightarrow\left\{{}\begin{matrix}x^{2k}=1\\x^{2k+1}=-1\end{matrix}\right.\) ta có:

\(Q=2017-2016+2015-2014+...+3-2+1\)

\(Q=1+1+1+...+1+1\) (có \(\frac{2016}{2}+1=1009\) số 1)

\(Q=1009\)

\(x^4+2017x^2+2016x+2017\)

\(=\left(x^4+x^2+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^4+2x^2+1-x^2\right)+2016\left(x^2+x+1\right)\)

\(=\left[\left(x^2+1\right)-x^2\right]+2016\left(x^2+x+1\right)\)

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

\(=\left(x^2+x+1\right)\left(x^2-x+2017\right)\)

\(x^4+2017x^2+2016x+2017\)

\(=\left(x^4-x\right)+\left(2007x^2+2007x+2007\right)\)

\(=x.\left(x^3-1\right)+2007.\left(x^2+x+1\right)\)

\(=x.\left(x-1\right)\left(x^2+x+1\right)+2007.\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2007\right)\)

cảm ơn các bạn! 

Ta có: \(\left|x+\frac{1}{2015}\right|\ge0\)

\(\left|x+\frac{2}{2015}\right|\ge0\)

...

\(\left|x+\frac{2016}{2015}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{2015}\right|+\left|x+\frac{2}{2015}\right|+...+\left|x+\frac{2016}{2015}\right|\ge0\)

\(\Rightarrow2017x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\frac{1}{2015}\right|+\left|x+\frac{2}{2015}\right|+...+\left|x+\frac{2016}{2015}\right|=x+\frac{1}{2015}+x+\frac{2}{2015}+...+x+\frac{2016}{2015}=2017x\)

\(\Rightarrow2016x+\left(\frac{1}{2015}+\frac{2}{2015}+...+\frac{2016}{2015}\right)=2017x\)

\(\Rightarrow x=\frac{1+2+...+2016}{2015}\)

Vậy \(x=\frac{1+2+...+2016}{2015}\)

Bạn cần số cụ thể thì tính ra nhé!

Đặt 2017x-2016=a; 2016x-2015=b

Theo đề, ta có: \(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

\(\Leftrightarrow x\in\left\{\dfrac{2016}{2017};\dfrac{2015}{2016};\dfrac{4031}{4033}\right\}\)

giúp mình vx

dễ vcl :)))

( 2017x₁ - 2016y₂ )² + ( 2017x₂ - 2016y² )² + ... + ( 2017x₂₀₁₆ - 2016x₂₀₁₆)² 

Chẳng có quy luật gì cả :)))

Hình như sai đề 

----

sửa đề: \(\left(2017x_1-2016y_1\right)^2+\left(2017x_2-2016y_2\right)^2+....+\left(2017x_{2016}-2016y_{2016}\right)^2\le0\)

\(\text{ Ta có: }VT\ge0\text{ mà }VP\le0\)

Dấu = xảy ra khi: \(\left(2017x_1-2016y_1\right)^2+\left(2017x_2-2016y_2\right)^2+....+\left(2017x_{2016}-2016y_{2016}\right)^2=0\)

\(\Rightarrow2017x_1-2016y_1=0;2017x_2-2016y_2=0;.....;2017x_{2016}-2016y_{2016}=0\)

\(\Rightarrow2017x_1=2016y_1;2017x_2=2016_{y2};.....\text{và }2017x_{2016}=2016y_{2016}\)

\(\Rightarrow x_1=\frac{2016y_1}{2017};x_2=\frac{2016y_2}{2017};....;x_{2016}=\frac{2016y_{2016}}{2017}\)

=> \(\frac{x_1+x_2+x_3+...+x_{2016}}{y_1+y_2+y_3+...+y_{2016}}=\frac{\frac{2016.\left(y_1+y_2+...+y_{2016}\right)}{2017}}{y_1+y_2+...+y_{2016}}=\frac{2016}{2017}\)(đpcm)

P(2016)= 2016⁴-2017.2016³+2017.2016²-2017.2016+2017

P(2016)=1

mk mới học lớp 5 lên ko bit

Ta có : x^4+2017x^2+2016x+2017

=x^4+x^3-x^3+x^2-x^2+2017x^2+2017x-x+2017

=x^4+x^3+x^2-x^3-x^2-x+2017x^2+2017x+2017

=x^2(x^2+x+1)-x(x^2+x+1)+2017(x^2+x+1)

=(x^2+x+1)(x^2-x+2017)

Nhớ k mk nha

Ta có : x^4+2017x^2+2016x+2017
=x^4+x^3-x^3+x^2-x^2+2017x^2+2017x-x+2017
=x^4+x^3+x^2-x^3-x^2-x+2017x^2+2017x+2017
=x^2(x^2+x+1)-x(x^2+x+1)+2017(x^2+x+1)
=(x^2+x+1)(x^2-x+2017)

chúc cậu hok tốt _@

2017 = 2016 + 1 = x + 1

suy ra 2017x¹⁵ = x¹⁶ + x¹⁵

2017x¹⁴ = x¹⁵ + x¹⁴

.... 

từ đó ta dễ tính ra A