• CCCCCCCCCCCChiuj

6x-2-5=2016x-2017

6x-7=2016x-2017

2016x-6x=2017-7

2010x=2010

x=1

Ta có: x=2017 suy ra: x+1=2018

Thay x+1=2018 vào biểu thức A,ta có:

\(x^{20}-x+1.x^{19}-x+1.x^{18}-...-x+1.x-1=x^{20}-x^{20}-x^{19}+x^{19}+...+x-1\)

Suy ra: x-1=2017-1=2016

Vậy x=2016

a) \(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{3}=-\frac{\sqrt{3}}{2}\\x-\sqrt{3}=\frac{\sqrt{3}}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{2}\\\frac{3\sqrt{3}}{2}\end{cases}}\)

Nghiệm cuối cùng là : \(x_1=\frac{\sqrt{3}}{2};x_2=\frac{3\sqrt{3}}{2}\)

b) || 6x - 2  | - 5 | = 2016. x -2017 

<=> || 6x - 2 | -5 | -2016x = -2017

<=> \(\orbr{\begin{cases}\left|6x-2\right|-5-2016.x=-2017,\left|6x-2\right|-5\ge0\\-\left(\left|6x-2\right|-5\right)-2016x=-2017,\left|6x-2\right|-5< 0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1,x\in\left[-\infty,-\frac{1}{2}\right];\left[\frac{7}{6};+\infty\right]\\x=\frac{1012}{1011},x\in\left[-\frac{1}{2},\frac{7}{6}\right]\end{cases}}\)

<=>\(\orbr{\begin{cases}x\in\varnothing\\x=\frac{1012}{1011}\end{cases}}\)

Vậy x = \(\frac{1012}{1011}\)

x=2017 nen x-1=2016

\(A=x^{10}-x^9\left(x-1\right)-x^8\left(x-1\right)-...-x\left(x-1\right)-1\)

\(=x^{10}-x^{10}+x^9-x^9+x^8-...-x^2+x-1\)

=x-1=2016