\(A=\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}=\sqrt[3]{\left(\sqrt{3}+1\right)^3}-\sqrt[3]{\left(\sqrt{3}-1\right)^3}=\sqrt{3}+1-\sqrt{3}+1=2\)

CHIM CÁNH CỤT 

TICK NHA

ta có:

\(X=\dfrac{\sqrt[3]{10+6\sqrt{3}}\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=2\)

thay x vào A, ta có A=1

\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{2\sqrt{x}+1}{x+\sqrt{x}-2}\)

\(A=\dfrac{\sqrt{x}+1+\sqrt{x}}{x-1}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)

\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

áp dụng hệ thức lượng trong tam giác vuông vào tam giác ABC vuông tại B, ta có:\(\left\{{}\begin{matrix}AB=cos\left(30\right).50=25\sqrt{3}\\BC=sin\left(30\right).50=25\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}P=\left(25+25\sqrt{3}\right).2\approx136,6\left(cm\right)\\S=25.25\sqrt{3}\approx1082,53\left(cm^2\right)\end{matrix}\right.\)

\(B=\left(\dfrac{5+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\right)^2-\left(\dfrac{5-2\sqrt{6}}{\sqrt{3}-\sqrt{2}}\right)^2\)

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

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

\(B=10\)