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Đặt \(\sqrt[3]{a}=x;\sqrt[3]{b}=y\)
=>\(Q=\dfrac{x^4+x^2y^2+y^4}{x^2+xy+y^2}\)
\(=\dfrac{x^4+2x^2y^2+y^4-x^2y^2}{x^2+xy+y^2}\)
\(=\dfrac{\left(x^2+y^2\right)^2-\left(xy\right)^2}{x^2+xy+y^2}=\dfrac{\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)}{x^2+xy+y^2}\)
\(=x^2-xy+y^2\)
\(=\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}\)
a/ \(\sqrt{8\left(\sqrt{2}-\sqrt{3}\right)^2}=2\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)=2\sqrt{6}-4\)
b/ \(ab\sqrt{1+\frac{1}{a^2b^2}}=ab.\sqrt{\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2.\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2+1}\)
c/ \(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}=\sqrt{\frac{a}{b^3}\left(1+\frac{1}{b}\right)}=\frac{1}{b}.\sqrt{\frac{a}{b}\left(1+\frac{1}{b}\right)}\)
d/ \(\frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{a}=x\\\sqrt[3]{b}=y\end{matrix}\right.\) thì ta có:
\(Q=\dfrac{x^4+x^2y^2+y^4}{x^2+xy+y^2}=\dfrac{\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)}{x^2+xy+y^2}=x^2-xy+y^2\)
Vậy \(Q=\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}\)
b: \(A=\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)
\(=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\)
\(\Leftrightarrow A^3=4+\sqrt{15}+4-\sqrt{15}+3\cdot A\cdot1\)
\(\Leftrightarrow A^3-3A-8=0\)
hay \(A\simeq2.49\)
a: \(B=\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\)
\(\Leftrightarrow B^3=5-\sqrt{17}+5+\sqrt{17}+3\cdot B\cdot2=10+6B\)
\(\Leftrightarrow B^3-6B-10=0\)
hay \(B\simeq3.05\)
\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)
\(=14\)
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
dễ quá đi mất
Phúc dễ thj lm đi