Sai cả rồi. Phải là 1909.79 mới đúng

\(x^2-3x+1=\dfrac{5\sqrt{3}}{3}\sqrt{x^4+x^2+1}\)

\(\Leftrightarrow\)\(\left(x^2-3x+1\right)^2=\dfrac{25}{3}\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\)\(x^4-6x^3+11x^2-6x+1=\dfrac{25}{3}x^4+\dfrac{25}{3}x^2+\dfrac{25}{3}\)

\(\Leftrightarrow11x^4+9x^3-4x^2+9x+11=0\)

\(\Leftrightarrow\left(x+1\right)\left(11x^3-2x^2-2x+11\right)=0\)

\(\Rightarrow x=-1\)

a/ \(A=\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}\)

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

\(A=\sqrt{3}+1-\sqrt{2}-\sqrt{3}+\sqrt{2}\)

\(A=1\)

b/ \(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)

\(B=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

\(B=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)

\(B=\sqrt{25}=5\)