x^2017+x^2015+1=(x^2017-x)+(x^2015-x^2)+(x^2+x+1) (1)

Ta có:x^2017-x=x(x^2016-1)

Dễ thấy x^2016-1 chia hết cho x^3-1 hay chia hết cho x^2+x+1 suy ra x^2017-x chia hết cho x^2+x+1 (2)

Tương tự x^2015-x^2 chia hết cho x^2+x+1 (3)

và x^2+x+1 chia hết cho x^2+x+1 (4)

Từ (1)(2)(3)(4) ta có (đpcm).

Giải:

a) \(x^2+5x=6\)

\(\Leftrightarrow x^2+5x-6=0\)

\(\Leftrightarrow x^2+6x-x-6=0\)

\(\Leftrightarrow\left(x^2+6x\right)-\left(x+6\right)=0\)

\(\Leftrightarrow x\left(x+6\right)-\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)

Vậy ...

b) \(x^2-2015x+2014=0\)

\(\Leftrightarrow x^2-2014x-x+2014=0\)

\(\Leftrightarrow\left(x^2-2014x\right)-\left(x-2014\right)=0\)

\(\Leftrightarrow x\left(x-2014\right)-\left(x-2014\right)=0\)

\(\Leftrightarrow\left(x-2014\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2014=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2014\\x=1\end{matrix}\right.\)

Vậy ...

Chúc bạn học tốt!

\(a,x^2+5x=6\)

\(\Rightarrow x^2+5x-6=0\)

\(\Rightarrow x^2+6x-x-6=0\)

\(\Rightarrow\left(x^2-6x\right)-\left(x+6\right)=0\)

\(\Rightarrow x\left(x+6\right)-\left(x+6\right)=0\)

\(\Rightarrow\left(x+6\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)

\(b,x^2+2015x+2014=0\)

\(\Rightarrow x^2+2015x+2015-1=0\)

\(\Rightarrow\left(x^2-1\right)+\left(2015x+2015\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)+2015\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x-1+2015\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x+2014\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+2014=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2014\end{matrix}\right.\)

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)

Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)

<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)

<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)

<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)

<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)

<=> \(x=2017\)

Vậy x = 2017

đúng thì

gõ phân số ra cho mk nhìn đc ko cậu

đây nè mấy nàng ơi. trả lời câu này nhé . làm ơn đi

\(\left(2x-1\right)^3-8\left(x-1\right)\left(x^2+x+1\right)+12x^2=2x+1\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8\left(x^3-1\right)+12x^2-2x-1=0\)

\(\Leftrightarrow4x+6=0\)

\(\Leftrightarrow2\left(2x+3\right)=0\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=\frac{-3}{2}\)

đề sai nhé

phải là

8. (x-1) . (x²+x+1)