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\(x^3=6+3x\sqrt[3]{\left(3+\sqrt{\frac{368}{27}}\right)\left(3-\sqrt{\frac{368}{27}}\right)}\Leftrightarrow x^3=6+3x.\sqrt[3]{9-\frac{368}{27}}\Leftrightarrow x^3+5x-6=0\)
Tự làm tiếp nha
a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\dfrac{\sqrt{9\left(5+3\sqrt{2}\right)}+\sqrt{9\left(5-3\sqrt{2}\right)}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}{3+\sqrt{2}-3+\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{3+\sqrt{2}+2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}+3-\sqrt{2}}{2\sqrt{2}}\)
\(=\dfrac{3\left[5+3\sqrt{2}+5-3\sqrt{2}+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}\right]}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)
\(=\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}=\dfrac{30+6\sqrt{7}-18-6\sqrt{7}}{6\sqrt{2}}=\dfrac{12}{6\sqrt{2}}\)
\(=\sqrt{2}\)
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\sqrt{5+3\sqrt{2}}+3\sqrt{5-3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}\right)}-\dfrac{\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}}{\sqrt{6+3\sqrt{2}}-\sqrt{6-4\sqrt{2}}}\\ =\dfrac{3\left(5+3\sqrt{2}+2\sqrt{25-18}+5-3\sqrt{2}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{\sqrt{4+2+4\sqrt{2}}+\sqrt{4+2-4\sqrt{2}}}{\sqrt{4+2+4\sqrt{2}}-\sqrt{4+2-4\sqrt{2}}}\\ =\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}}{\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{2+\sqrt{2}+2-\sqrt{2}}{2+\sqrt{2}-2+\sqrt{2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{4}{2\sqrt{2}}=\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}=\dfrac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}=\dfrac{4+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}}{2\sqrt{2}}\) \(=\dfrac{4+2\sqrt{7}-2\sqrt{7}}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
Lời giải:
Đặt \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}=a; \sqrt[3]{2-10\sqrt{\frac{1}{27}}}=b\)
Khi đó: \(D=a+b\)
\(\left\{\begin{matrix} a^3+b^3=2+10\sqrt{\frac{1}{27}}+2-10\sqrt{\frac{1}{27}}=4\\ ab=\sqrt[3]{(2+10\sqrt{\frac{1}{27}})(2-10\sqrt{\frac{1}{27}})}=\frac{2}{3}\end{matrix}\right.\)
Có:
\(a^3+b^3=(a+b)^3-3ab(a+b)\)
\(\Leftrightarrow 4=D^3-3.\frac{2}{3}D\)
\(\Leftrightarrow D^3-2D-4=0\)
\(\Leftrightarrow (D-2)(D^2+2D+2)=0\)
Vì \(D^2+2D+2=(D+1)^2+1\neq 0\), do đó \(D-2=0\Leftrightarrow D=2\)
\(A=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)
\(\Leftrightarrow A=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)
\(\Leftrightarrow A^3=6+3A.\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}.\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)
\(\Leftrightarrow A^3=6+3A.\left(-\dfrac{5}{3}\right)\)
\(\Leftrightarrow A^3+5A-6=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+6\right)=0\)
\(\Leftrightarrow A=1\)
\(A=\sqrt[3]{3+\sqrt{\dfrac{368}{27}}}+\sqrt[3]{3-\sqrt{\dfrac{368}{27}}}\)
\(\Leftrightarrow A^3=6+3A.\sqrt[3]{-\dfrac{125}{27}}=6-5A\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+6\right)=0\)
Vì \(A^2+A+6>0\)
\(\Rightarrow A=1\)