bn ơi, đề bài của bạn chụp thiếu rùi

`(2x-3)(x^2+x+1)-x(2x^2-x-1)`

`=2x^3+2x^2+2x-3x^2-3x-3-2x^3+x^2+x`

`=(2x^3-2x^3)+(2x^2-3x^2+x^2)+(2x-3x+x)-3`

`=-3`

Trả lời:

( 2x - 3 ) ( x² + x + 1 ) - x ( 2x² - x - 1 )

= 2x³ + 2x² + 2x - 3x² - 3x - 3 - 2x³ + x² + x

= - 3

(x + 1)² - (2x - 1)² = 0

<=> (x + 1 + 2x - 1) (x + 1 - 2x + 1) = 0

<=> 3x (- x + 2) = 0

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\-x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy tập nghiệm pt: S = {0 ; 2}.

( x + 1 )² - ( 2x - 1 )² = 0

=> ( x + 1 )² = ( 2x - 1 )²

=> x + 1 = 2x - 1

=> x + 2 = 2x

=> 2x - x = 2

=> x = 2

Vậy x = 2

Đặt \(f\left(x\right)=\left(x+1\right)P\left(x\right)-x\).

Khi đó \(f\left(k\right)=0\)với mọi \(k=0,1,2,...,2018\)mà \(P\left(x\right)\)có bậc \(2018\)nên \(f\left(x\right)\)có bậc \(2019\)

mà \(f\left(x\right)=0\)tại \(2019\)giá trị nên \(f\left(x\right)=ax\left(x-1\right)\left(x-2\right)...\left(x-2018\right)\).

Với \(x=-1\): \(a.\left(-1\right)\left(-2\right)...\left(-2019\right)=\left(-1+1\right)P\left(-1\right)-\left(-1\right)\)

\(\Leftrightarrow a=-\frac{1}{2019!}\).

\(P\left(2019\right)=\frac{f\left(2019\right)+2019}{2020}=\frac{-1+2019}{2020}=\frac{1009}{1010}\)

Tổng các hệ số phi khai triển đa thức \(P\left(x\right)\)là \(P\left(1\right)\).

\(P\left(1\right)=\left(1^3-2.1^2+2\right)^{2018}=1^{2018}=1\)

Đa thức \(P\left(x\right)=x^3-3x+1\)có ba nghiệm phân biệt \(x_1,x_2,x_3\) có: 

\(\hept{\begin{cases}x_1+x_2+x_3=0\\x_1x_2+x_2x_3+x_3x_1=-3\\x_1x_2x_3=-1\end{cases}}\)

\(E=Q\left(x_1\right)Q\left(x_2\right)Q\left(x_3\right)=\left(x_1^2-1\right)\left(x_2^2-1\right)\left(x_3^2-1\right)\)

\(=\left(x_1x_2x_3\right)^2-\left(x_1^2x_2^2+x_2^2x_3^2+x_3^2x_1^2\right)+\left(x_1^2+x_2^2+x_3^2\right)-1\)

\(=\left(x_1x_2x_3\right)^2-\left[\left(x_1x_2+x_2x_3+x_3x_1\right)^2-2x_1x_2x_3\left(x_1+x_2+x_3\right)\right]+\left[\left(x_1+x_2+x_3\right)^2-2\left(x_1x_2+x_2x_3+x_3x_1\right)\right]-1\)

\(=\left(-1\right)^2-3^2+2.3-1=-3\)