Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}=\dfrac{1}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}\)
Đặt \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)\(\Leftrightarrow A^3=18+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow A^3=18+3A\sqrt[3]{1}\\ \Leftrightarrow A^3-3A-18=0\\ \Leftrightarrow A=3\\ \Leftrightarrow X=\dfrac{1}{3}\\ \Leftrightarrow Q=\left[3\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^2-1\right]^{2021}=\left(\dfrac{1}{9}-\dfrac{1}{9}-1\right)^{2021}=\left(-1\right)^{2021}=-1\)
\(A^2=12-\sqrt{80-32\sqrt{3}}+12+\sqrt{80-32\sqrt{3}}-2\sqrt{144-80+32\sqrt{3}}\)
=>\(A^2=24-2\sqrt{48+32\sqrt{3}}\)
=>A^2=24-8căn 3+2căn 3
=>\(A=\sqrt{24-8\sqrt{3+2\sqrt{3}}}\)
Đặt \(A=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow A^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+3.\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)^2\left(1-\frac{\sqrt{84}}{9}\right)}+3.\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)\left(1-\frac{\sqrt{84}}{9}\right)^2}\)
\(A^3=2+3.\sqrt[3]{-\frac{1}{27}.\left(1+\frac{\sqrt{84}}{9}\right)}+3.\sqrt[3]{-\frac{1}{27}.\left(1-\frac{\sqrt{84}}{9}\right)}\)
\(=2-\left(\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)}+\sqrt[.3]{\left(1-\frac{\sqrt{84}}{9}\right)}\right)\)
\(A^3=2-A\Leftrightarrow\left(A-1\right)\left(A^2+A+2\right)=0\Rightarrow A=1\)
Đặt \(A=\sqrt[3]{\frac{9+2\sqrt{21}}{9}}+\sqrt[3]{\frac{9-2\sqrt{21}}{9}}\)
\(A^3=\frac{9+2\sqrt{21}+9-2\sqrt{21}}{9}+3\sqrt[3]{\frac{9^2-4\cdot21}{9^2}}A\)
\(A^3-2+A=0\Leftrightarrow\left(A-1\right)\left(A^2+A+1\right)+A-1=0\Leftrightarrow\left(A-1\right)\left(A^2+A+2\right)=0\)
\(\Rightarrow A=1\)(ĐPCM)
Ta có \(\sqrt[3]{26+15\sqrt{3}}=\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}\)
\(=\sqrt[3]{2^3+3.2^2\sqrt{3}+3.2.\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}\)
Đặt \(x=\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)
Ta có \(x^3=\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)^3\)
\(=9+\sqrt{80}+9-\sqrt{80}+3.\left(\sqrt[3]{9+\sqrt{80}}\right)^2\left(\sqrt[3]{9-\sqrt{80}}\right)+3.\left(\sqrt[3]{9-\sqrt{80}}\right)^2\left(\sqrt[3]{9+\sqrt{80}}\right)\)
\(=18+3\sqrt[3]{9+\sqrt{80}}.\sqrt[3]{9-\sqrt{80}}\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)\)
\(=18+3\sqrt[3]{9^2-80}.x\)
\(=18+3x\)
Vậy \(x^3=18+3x\)
\(\Leftrightarrow x^3-3x-18=0\)
Vậy x = 3
Do đó \(M=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+3=2^2-3+3=4\)
Vậy M = 4.
\(A=\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)
\(A^3=\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)^3\)
\(=9+\sqrt{80}+9-\sqrt{80}+3\sqrt[3]{9+\sqrt{80}}.\sqrt[3]{9-\sqrt{80}}\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)\)\(=18+3.A\)
<=> \(A^3-3A-18=0\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)
<=> A=3
vì \(A^2+3A+6=\left(A+\frac{3}{2}\right)^2+\frac{15}{4}>0\)
Vậy A là một số nguyên