x=2018 nên x-1=2017

\(A=x^{10}-x^9\left(x-1\right)-x^8\left(x-1\right)-...-x^2\left(x-1\right)-x\left(x-1\right)-1\)

\(=x^{10}-x^{10}+x^9-x^9+x^8-...-x^3+x^2-x^2+x-1\)

=x-1=2017

\(A=x^3+2x^2+3x\\ =x\left(x^2+2x+1\right)\\ =x\left(x+1\right)^2\\ =1999.\left(1999+1\right)=1999.2000\\ =3998000\)

\(B=x^4-2017x^3+2017x^2-2017x+2018\\ =x^4-2016x^3-x^3+2016x^2+x^2-2016x-x+2016+2\\ =x^3\left(x-2016\right)-x^3\left(x-2016\right)+x\left(x-2016\right)-\left(x-2016\right)+2\\ =\left(x-2016\right)\left(x^3+x-1\right)+2=0+2=0\)

Bạn xem lại đề câu a nhé , theo mk thì phải là 2 thì tính ms nhanh đc, 3 thì cũng giải đc nhưng ko hợp lí lắm

\(A=\frac{1}{2017}-\frac{2}{2017x}+\frac{1}{x^2}=\left(\frac{1}{2017}-\frac{1}{x}\right)^2+\frac{1}{2017}-\frac{1}{2017^2}=\left(\frac{1}{2017}-\frac{1}{x}\right)^2+\frac{2016}{2017^2}\)

\(\Rightarrow A\ge\frac{2016}{2017^2}\)Dấu "=" xảy ra khi \(\left(\frac{1}{2017}-\frac{1}{x}\right)^2=0\Rightarrow x=2017\)

Vây ......

Câu 2 :

Ta có : \(x^3+2x^2-x-14=0\)

=> \(x^3-2x^2+4x^2-8x+7x-14=0\)

=> \(x^2\left(x-2\right)+4x\left(x-2\right)+7\left(x-2\right)=0\)

=> \(\left(x-2\right)\left(x^2+4x+7\right)=0\)

=> \(\left[{}\begin{matrix}x-2=0\\x^2+4x+7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=2\\x^2+4x+4+3=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=2\\\left(x+2\right)^2=-3\left(VL\right)\end{matrix}\right.\)

=> \(x=2\)

- Ta có : \(2x^4+5x^3-29x+80\)

\(=2x^4-4x^3+9x^3-18x^2+18x^2-36x+7x-14+94\)

\(=2x^3\left(x-2\right)+9x^2\left(x-2\right)+18x\left(x-2\right)+7\left(x-2\right)+94\)

\(=\left(2x^3+9x^2+18x+7\right)\left(x-2\right)+94\left(I\right)\)

- Thay x = 2 vào biểu thức ( I ) ta được :

\(\left(2.2^3+9.2^2+18.2+7\right)\left(2-2\right)+94\)

\(=\left(2.2^3+9.2^2+18.2+7\right)0+94\)

\(=0+94\)

\(=94\)

Vậy giá trị của biểu thức trên là 94 .

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)

\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)

Ta có : x - 1 = 2018 - 1 = 2017 

N = x⁶ - 2017x⁵ - 2017x⁴ - 2017x³ - 2017x² - 2017x - 2017

N = x⁶ - ( x - 1 ).x⁵ - ( x - 1 ).x⁴ - ( x - 1 ).x³ - ( x - 1 ).x² - ( x - 1 ).x - ( x - 1 )

N = x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x + 1

N = 1

\(a,ĐK:x\ne1;x\ne-1\\ b,C=\dfrac{x^2+x+x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+2x+1}{2x^2-2}\\ c,C=-\dfrac{1}{2}\Leftrightarrow2-2x^2=2x^2+2x+1\\ \Leftrightarrow4x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-1}{4}\\x=\dfrac{-\sqrt{5}-1}{4}\end{matrix}\right.\\ d,C>0\Leftrightarrow2x^2-2>0\left(2x^2+2x+1>0\right)\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)