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Ta có:
\(x=\frac{2}{2\sqrt[3]{2}+2\sqrt[3]{4}}=\frac{1}{\sqrt[3]{2}+\sqrt[3]{4}}=\)\(\frac{2\sqrt[3]{2}-2+\sqrt[3]{4}}{6}\)
\(y=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{6\left(\sqrt[3]{2}+\sqrt[3]{4}\right)}{6}\)
\(\Rightarrow xy^3-x^3y=xy\left(y^2-x^2\right)=y^2-x^2=\frac{36\left(\sqrt[3]{4}+4+2\sqrt[3]{2}\right)}{36}\)\(-\frac{4\sqrt[3]{4}+4+2\sqrt[3]{2}-8\sqrt[3]{2}+8-4\sqrt[3]{4}}{36}\)\(=\frac{36\sqrt[3]{4}+144+72\sqrt[3]{2}-12+6\sqrt[3]{2}}{36}=\frac{36\sqrt[3]{4}+78\sqrt[3]{2}+132}{36}\)\(=\frac{6\sqrt[3]{4}+13\sqrt[3]{2}+22}{6}\)
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: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(3+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{9-3}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{3+\sqrt{3}+3-\sqrt{3}}{6}=\dfrac{6}{6}=1\)
\(x=\sqrt{2}\)
\(y=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(y\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)
\(y\sqrt{2}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(y\sqrt{2}=\sqrt{7}+1-\sqrt{7}+1\)
\(y\sqrt{2}=2\)
\(y=\dfrac{2}{\sqrt{2}}\)
Thay \(x=\sqrt{2},y=\dfrac{2}{\sqrt{2}}\) vào A ta có:
\(A=\dfrac{\sqrt{2}.\dfrac{2}{\sqrt{2}}-1}{\sqrt{2}+\dfrac{2}{\sqrt{2}}}-\dfrac{1-\sqrt{2}.\dfrac{2}{\sqrt{2}}}{2\sqrt{2}-\dfrac{2}{\sqrt{2}}}\)
\(=\dfrac{2-1}{2\sqrt{2}}-\dfrac{1-2}{\sqrt{2}}\)
\(=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{4}\)
Tự kết luận nha
Áp dụng BĐT AM-GM, Ta có
\(\sqrt{x-1}\le\dfrac{1+x-1}{2}=\dfrac{x}{2}\Rightarrow yz\sqrt{x-1}\le\dfrac{xyz}{2}\)
Mà \(xz\sqrt{y-2}\le\dfrac{xz\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\)
\(yx\sqrt{z-3}\le yx.\dfrac{3+z-3}{2\sqrt{3}}=\dfrac{xyz}{2\sqrt{3}}\)
\(\Rightarrow\dfrac{xy\sqrt{x-1}+xz\sqrt{y-2}+yz\sqrt{z-3}}{xyz}\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}=\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{3}}{6}\)