Ta có: x=2017

nên x+1=2018

Ta có: \(P=x^{15}-2018x^{14}+2018x^{13}-2018x^{12}+...+2018x^3-2018x^2+2018x-2018\)

\(=x^{15}-\left(x+1\right)\cdot x^{14}+\left(x+1\right)\cdot x^{13}-\left(x+1\right)\cdot x^{12}+...+\left(x+1\right)\cdot x^3-\left(x+1\right)\cdot x^2+\left(x+1\right)\cdot x-\left(x+1\right)\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}+...+x^3-x^3+x^2-x^2+x-x-1\)

=-1

@ 肖战Daytoy_1005 giup

Đề:............

<=> - (1 - 2018x) + 2019x.(1 - 2018x) = 0

<=> (1 - 2018x).[(-1) + 2019x] = 0

Xét 2 trường hợp, ta có:

TH₁: 1 - 2018x = 0          TH₂: -1 + 2019x = 0

<=> 2018x = 1                 <=> 2019x = 1

<=> x = 1/2018                <=> x = 1/2019

Vậy x = 1/2018; 1/2019

\(2018x-1+2019x\left(1-2018x\right)=0\)

\(-\left(1-2018x\right)+2019x\left(1-2018x\right)=0\)

\(\left(1-2018x\right)\left(-1+2019x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}1-2018x=0\\-1+2019x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2018}\\x=\frac{1}{2019}\end{cases}}}\)

F(x)=\(x^7-2018x^6+2018x^5-2018x^4+2018x^3-2018x^2+2018x+1.\)

x=2017=>2018=x+1 thay vào F(x) ta có:

F(x)=x+1=2018

pkm;lkml

\(A=x^9-2018x^8+2018x^7-2018x^6+2016x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)

\(A=x^9-\left(2017+1\right)x^8+\left(2017+1\right)x^7-...+\left(2017+1\right)x-\left(2017+1\right)\)

\(A=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-...+\left(x+1\right)x-x-1\)

\(A=x^9-x^9-x^8+x^8+x^7-...+x^2+x-x-1\)

\(A=-1\)

x² - 5x = 0

=> x(x - 5) = 0

=> \(\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

b)  (3x - 5)² - 4 = 0

=> (3x - 5)² = 0 + 4

=> (3x - 5)² = 4

=> (3x - 5)² = 2²

=> \(\orbr{\begin{cases}3x-5=2\\3x-5=-2\end{cases}}\)

=> \(\orbr{\begin{cases}3x=7\\3x=3\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)

\(2018x^2-2019x+1=0\)

\(2018x^2-2018x-x+1=0\)

\(2018x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(x-1\right)\left(2018x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2018x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2018}\end{cases}}}\)

= \(\frac{1}{2018}\)

\(a,\Leftrightarrow2x^2-10x-2x^2-x=-11\\ \Leftrightarrow-11x=-11\Leftrightarrow x=1\\ b,\Leftrightarrow x\left(x^2-6x+9\right)=0\\ \Leftrightarrow x\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-2018\right)-2017\left(x-2018\right)=0\\ \Leftrightarrow\left(x-2017\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2018\end{matrix}\right.\)

Vì \(x=2017\Rightarrow x+1=2018\)

Thay \(x+1=2018\)vào biểu thức A ta được :

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

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

\(=1\)

Tại x=2017 thì 2018 = x + 1 

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

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

\(=1\)