$\displaystyle \begin{array}{{>{\displaystyle}l}} 54\\ Cho\ a=\sqrt[3]{4} +\sqrt[3]{2} +1.\ Tính\ \frac{3}{a} +\frac{3}{a^{2}} +\frac{1}{a^{3}}\\ \frac{3}{a} +\frac{3}{a^{2}} +\frac{1}{a^{3}} =\frac{1}{a}\left( 3+\frac{3}{a} +\frac{1}{a^{2}}\right)\\ =\frac{1}{a}\left(\frac{1}{a^{2}} +\frac{3}{a} +\frac{9}{4} +\frac{3}{4}\right) =\frac{1}{a}\left(\left(\frac{1}{a} +\frac{3}{2}\right)^{2} +\frac{3}{4}\right) \ \\ a=\left(\sqrt[3]{2}\right)^{2} +\sqrt[3]{2} +1\\ \rightarrow a\left(\sqrt[3]{2} -1\right) =3\ \\ \rightarrow a=\frac{3}{\sqrt[3]{2} -1} \ \ \\ Thay\ vào\ tính\ nha\ \\ \\ Bài\ :52\sqrt{1+x} +\sqrt{1+y} =2\sqrt{1+a} \ \\ ( x+y)^{2} \geqslant 2\left( x^{2} +y^{2}\right) \ \\ \rightarrow \left(\sqrt{1+x} +\sqrt{1+y}\right)^{2} \leqslant 2( 2+x+y) \ \\ \rightarrow 4( 1+a) \leqslant 4+2( x+y) \ \\ \rightarrow 4a\leqslant 2( x+y) \ \\ \rightarrow 2a\leqslant x+y\ \rightarrow dpcm\ \\ \\ \\ \\ \ \\ \ \\ \\ \end{array}$
