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\(x=\dfrac{3}{\sqrt[3]{4}-\sqrt[3]{2}+1}=\dfrac{3\left(\sqrt[3]{2}+1\right)}{\left(\sqrt[3]{2}+1\right)\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)}=\dfrac{3\left(\sqrt[3]{2}+1\right)}{3}=3\sqrt[3]{2}+1\)
\(y=\dfrac{6}{\sqrt[3]{16}+\sqrt[3]{4}+4}=\dfrac{6}{\sqrt[3]{4}\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)}=\dfrac{6\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}\left(\sqrt[3]{4}-1\right)\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)}=\dfrac{6\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}.3}=\dfrac{2\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}}=\dfrac{2\sqrt[3]{4}}{\sqrt[3]{4}}-\dfrac{\sqrt[3]{8}}{\sqrt[3]{4}}=2-\sqrt[3]{2}\)
=> x + y = \(\sqrt[3]{2}+1+2-\sqrt[3]{2}=3\)
a: \(=\dfrac{2\sqrt{7}-10-6+\sqrt{7}}{4}+\dfrac{24+6\sqrt{7}-20+5\sqrt{7}}{9}\)
\(=\dfrac{3\sqrt{7}-16}{4}+\dfrac{4+11\sqrt{7}}{9}\)
\(=\dfrac{27\sqrt{7}-144+16+44\sqrt{7}}{36}=\dfrac{71\sqrt{7}-128}{36}\)
b: \(=\dfrac{\sqrt{y}\left(x+y\right)}{\sqrt{xy}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{x+y}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}\)
c: \(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)+3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right)\cdot\dfrac{3\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+3\sqrt{x}-1}{3\sqrt{x}+1}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-5\right)}\)
Ta có: +) \(3=\left(\sqrt[3]{2}\right)^3+1^3=\left(\sqrt[3]{2}+1\right)\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)\Rightarrow\frac{1}{\sqrt[3]{4}-\sqrt[3]{2}+1}=\frac{\sqrt[3]{2}+1}{3}\)\(\Rightarrow\frac{3}{\sqrt[3]{4}-\sqrt[3]{2}+1}=\sqrt[3]{2}+1\)hay \(x=\sqrt[3]{2}+1\)
+) \(3=\left(\sqrt[3]{4}\right)^3-1^3=\left(\sqrt[3]{4}-1\right)\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)\)\(\Rightarrow\sqrt[3]{16}+\sqrt[3]{4}+1=\frac{3}{\sqrt[3]{4}-1}\Rightarrow4+\sqrt[3]{4}+\sqrt[3]{16}=\frac{3\sqrt[3]{4}}{\sqrt[3]{4}-1}\)\(\Rightarrow\frac{6}{4+\sqrt[3]{4}+\sqrt[3]{16}}=\frac{6\sqrt[3]{4}-6}{3\sqrt[3]{4}}=2-\frac{2}{\sqrt[3]{4}}=2-\sqrt[3]{2}\)hay \(y=2-\sqrt[3]{2}\)
Từ đó suy ra \(x+y=\sqrt[3]{2}+1+2-\sqrt[3]{2}=3\)là một số tự nhiên (đpcm)
Ta có: \(x=\frac{3\left(1+\sqrt[2]{2}\right)}{\left(\sqrt[3]{2^2}-\sqrt[3]{2}+1\right)\left(1+\sqrt[3]{2}\right)}=\frac{3\left(1+\sqrt[2]{2}\right)}{1+\left(\sqrt[3]{2}\right)^3}=1+\sqrt[2]{2}\)
\(y=\frac{6\left(2-\sqrt[3]{2}\right)}{\left(2^2+2\sqrt[3]{2}+\sqrt[3]{2^2}\right)\left(2-\sqrt[3]{2}\right)}=\frac{6\left(2-\sqrt[3]{2}\right)}{2^3-\left(\sqrt[3]{2}\right)^3}=2-\sqrt[3]{2}\)
Vậy x+y=1+\(\sqrt[3]{2}+2-\sqrt[3]{2}=3\)là 1 số tự
nhiên
a: \(A=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}=10\)
b: \(B=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}=-2\sqrt{y}\)
c: \(C=\dfrac{\sqrt{3}-1}{\sqrt{6}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)