Ta có: \(x=2-\sqrt{3}\)\(\Rightarrow2-x=\sqrt{3}\)\(\Rightarrow\left(2-x\right)^2=3\)\(\Rightarrow4-4x+x^2=3\)\(\Rightarrow x^2-4x+1=0\)

Lại có: \(B=x^5-3x^4-3x^3+6x^2-20x+2018\)

\(\Rightarrow B=x^5-4x^4+x^4+x^3-4x^3+5x^2+x^2+20x+5+2013\)

\(\Rightarrow B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2013\)

\(\Rightarrow B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2013\)

\(\Rightarrow B=x^3\cdot0+x^2\cdot0+5\cdot0+2013=2013\)

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

Dễ thấy x là nghiệm của PT \(x^2-4x+1\)

\(H=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2019\\ H=\left(x^2-4x+1\right)\left(x^3+x^2+5\right)+2019\\ H=2019\)

b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)

\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)

\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)

\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)

Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)

\(x+\sqrt{3}=2\Rightarrow\sqrt{3}=2-x\Rightarrow3=\left(2-x\right)^2\Rightarrow x^2-4x+1=0\)

Ta có:

\(B=x^5-4x^4+x^4-4x^3+x^3+5x^2+x^2-20x+5+2013\)

\(\Rightarrow B=x^5-4x^4+x^3+x^4-4x^3+x^2+5x^2-20x+5+2013\)

\(\Rightarrow B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2013\)

\(\Rightarrow B=x^3.0+x^2.0+5.0+2013=2013\)

Với giả thiết x² - 4x + 1 = 0 thì\(B=x^5-3x^4-3x^3+6x^2-20x+2025=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2020=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2020=\left(x^3+x^2+5\right)\left(x^2-4x+1\right)+2020=2020\)

Thank you nhiều nha . Chúc bạn học tốt. I love you <3

Bạn có thể giải tiếp cho mình ko vậy

\(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\)

\(\Leftrightarrow x^2-4x+1=0\)

ta có : \(P=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2013=2013\)