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

\(B=x^5-3x^4-3x^3+6x^2-20x+2022\)

\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2017\)

\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2017\)

\(B=2017\)(Do \(x^2-4x+1=0\))

ĐS: ...

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

\(B=x^5-3x^4-3x^3+6x^2-20x-2022\)

\(=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+5\left(x^2-4x+1\right)+2017\)

\(=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2017\)

\(=2017\)

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Đặt \(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\)

\(=6+2\sqrt{9-\left(5+2\sqrt{3}\right)}=6+2\sqrt{3+2\sqrt{3}+1}\)

\(=6+2\left(3+1\right)=6+6+2=14\)

Nên biểu thức tương đương với \(14-\sqrt{3}\)

Đặt \(\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}=P\)

\(\Rightarrow P^2=3+\sqrt{5+2\sqrt{3}}+3-\sqrt{5+2\sqrt{3}}+2\sqrt{9-\left(5+2\sqrt{3}\right)}\)

\(=6+2\sqrt{\left(\sqrt{3}-1\right)^2}=6+2\sqrt{3}-2=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)

\(\Rightarrow P-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)

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

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

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

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: \(B=x^5-3x^4-3x^3+6x^2-20x+2018\)

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

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

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

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

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

Ta có:

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

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

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

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

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

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

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