\(Q=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

\(=\sqrt[3]{8+12\sqrt{2}+12+2\sqrt{2}}+\sqrt[3]{8-12\sqrt{2}+12-2\sqrt{2}}\)

\(=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)

\(=2+\sqrt{2}+2-\sqrt{2}=4\)

Làm tiếp nhé

a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

\(A=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{2\sqrt{2}+3.2.2+3.\sqrt{2}.4+8}+\sqrt[3]{8-3.\sqrt{2}.4+3.2.2-2\sqrt{2}}=\sqrt[3]{\left(\sqrt{2}+2\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}=\sqrt{2}+2+2-\sqrt{2}=4=\sqrt{16}\) \(B=2\sqrt{5}=\sqrt{20}\)

⇒ \(A< B\)

\(A=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

\(A^3=20+14\sqrt{2}+20-14\sqrt{2}+3\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right).\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}\)\(A^3=40+6A\)

\(\Leftrightarrow A^3-6A-40=0\)

\(\Leftrightarrow A^3-4A^2+4A^2-16A+10A-40=0\)

\(\Leftrightarrow A^2\left(A-4\right)+4A\left(A-4\right)+10\left(A-4\right)=0\)

\(\Leftrightarrow\left(A-4\right)\left(A^2+4A+10\right)=0\)

Do: \(A^2+4A+10=\left(A+2\right)^2+6\)

\(\Leftrightarrow A=4=\sqrt{16}< B=2\sqrt{5}=\sqrt{20}\)

\(\Rightarrow A< B\)

a: \(\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\)

\(\left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

mà \(-2\sqrt{105}>-2\sqrt{120}\)

nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

b: \(\left(\sqrt{2}+\sqrt{8}\right)^2=10+2\cdot4=16=12+4\)

\(\left(3+\sqrt{3}\right)^2=12+6\sqrt{3}\)

mà \(4< 6\sqrt{3}\)

nên \(\sqrt{2}+\sqrt{8}< 3+\sqrt{3}\)

\(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\Leftrightarrow A^3=4+3\sqrt[3]{-1}.A\Leftrightarrow A^3=4-3A\Leftrightarrow A^3+3A-4=0\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)(1)

Ta có \(A^2+A+4>0\)

Vậy (1)\(\Leftrightarrow A-1=0\Leftrightarrow A=1\)

Vậy A=1

\(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\Leftrightarrow B^3=5\sqrt{2}+7-5\sqrt{2}+7-3\sqrt[3]{\left(5\sqrt{2}+7\right)\left(5\sqrt{2}-7\right)}\left(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\right)\Leftrightarrow B^3=14-3\sqrt[3]{1}.B\Leftrightarrow B^3=14-3B\Leftrightarrow B^3+3B-14=0\Leftrightarrow\left(B-2\right)\left(B^2+2B+7\right)=0\left(2\right)\)

Ta lại có \(B^2+2B+7>0\)

Vậy (2)\(\Leftrightarrow B-2=0\Leftrightarrow B=2\)

Vậy B=2

\(C=\sqrt[3]{20+14\sqrt{2}}-\sqrt[3]{14\sqrt{2}-20}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.\left(\sqrt{2}\right)^2.2+3.\sqrt{2}.4+8}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.\left(\sqrt{2}\right)^2.2+3.\sqrt{2}.4-8}=\sqrt[3]{\left(\sqrt{2}+2\right)^2}-\sqrt[3]{\left(\sqrt{2}-2\right)}=\sqrt{2}+2-\sqrt{2}+2=4\)

a: \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}+3\cdot A\)

=>A^3-3A-18=0

=>A=3

b: \(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

=>\(B^3=5\sqrt{2}+7-5\sqrt{2}+7+3B\)

=>B^3-3B-14=0

=>B=2,82

c: \(C^3=20+14\sqrt{2}-14\sqrt{2}+20-6C\)

=>C^3+6C-40=0

=>C=2,84

Câu b, c tương tự câu a. Mình làm câu a coi như tượng trưng nha !!!!!!

a) Đặt: \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

<=> \(A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}.\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

<=> \(A^3=4+3\sqrt[3]{4-5}.A\)

<=> \(A^3=4-3A\)

<=> \(A^3+3A-4=0\)

<=> \(\left(A-1\right)\left(A^2+A+4\right)=0\)

Có:     \(A^2+A+4=\left(A+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

=>    \(A-1=0\)

<=> \(A=1\)

=> \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1\)

VẬY TA CÓ ĐPCM