Trả lời:

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

\(2E=2.\sqrt[3]{\sqrt{5}-2}+2.\sqrt[3]{\sqrt{5}+2}\)

\(2E=\sqrt[3]{8\sqrt{5}-16}+\sqrt[3]{8\sqrt{5}+16}\)

\(2E=\sqrt[3]{5\sqrt{5}-15+3\sqrt{5}-1}+\sqrt[3]{5\sqrt{5}+15+3\sqrt{5}+1}\)

\(2E=\sqrt[3]{\left(\sqrt{5}-1\right)^3}+\sqrt[3]{\left(\sqrt{5}+1\right)^3}\)

\(2E=\sqrt{5}-1+\sqrt{5}+1\)

\(2E=2\sqrt{5}\)

\(E=\sqrt{5}\)

\(F=\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\)

\(F=\sqrt[3]{182+25\sqrt{53}}+\sqrt[3]{182-25\sqrt{53}}\)

\(2F=2.\sqrt[3]{182+25\sqrt{53}}+2.\sqrt[3]{182-25\sqrt{53}}\)

\(2F=\sqrt[3]{1456+200\sqrt{53}}+\sqrt[3]{1456-200\sqrt{53}}\)

\(2F=\sqrt[3]{343+147\sqrt{53}+1113+53\sqrt{53}}+\sqrt[3]{343-147\sqrt{53}+1113-53\sqrt{53}}\)

\(2F=\sqrt[3]{\left(7+\sqrt{53}\right)^3}+\sqrt[3]{\left(7-\sqrt{53}\right)^3}\)

\(2F=7+\sqrt{53}+7-\sqrt{53}\)

\(2F=14\)

\(F=7\)

Đặt \(x=\sqrt[3]{182+\sqrt[]{33125}}+\sqrt[3]{182-\sqrt[]{33125}}\)

\(\Rightarrow x^3=364+3\sqrt[3]{182^2-33125}.\left(\sqrt[3]{182+\sqrt[]{33125}}+\sqrt[]{182-\sqrt[]{33125}}\right)\)

\(\Rightarrow x^3=364+3.\left(-1\right).x\)

\(\Rightarrow x^3+3x-364=0\)

\(\Rightarrow\left(x-7\right)\left(x^2+7x+52\right)=0\)

\(\Rightarrow x-7=0\) (do \(x^2+7x+52>0;\forall x\))

\(\Rightarrow x=7\)

a/ \(D\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\Rightarrow D=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

b/\(2E=\sqrt[3]{8\sqrt{5}-16}+\sqrt[3]{8\sqrt{5}+16}\)

\(=\sqrt[3]{5\sqrt{5}-3.5.1+3\sqrt{5}-1}+\sqrt[3]{5\sqrt{5}+3.5.1+3\sqrt{5}+1}\)

\(=\sqrt[3]{\left(\sqrt{5}-1\right)^3}+\sqrt[3]{\left(\sqrt{5}+1\right)^3}=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)

\(\Rightarrow E=\sqrt{5}\)

c/

\(F=\sqrt[3]{182+25\sqrt{53}}+\sqrt[3]{182-25\sqrt{53}}\)

\(F^3=364+3F\sqrt[3]{182^2-33125}=364-3F\)

\(\Leftrightarrow F^3+3F-364=0\)

\(\Leftrightarrow\left(F-7\right)\left(F^2+7F+52\right)=0\)

\(\Rightarrow F=7\)

Bài 2:

a/ \(C=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right)\left(\sqrt{4}+\sqrt{3}\right)}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}\)

\(=\sqrt{4}-1=2-1=1\)

Bài 2

b/

\(D=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)

\(=\sqrt{2}\)

c/

\(E=\frac{\left(\sqrt{3}-x\right)\left(\sqrt{3}+x\right)}{x+\sqrt{3}}=\sqrt{3}-x\)

d/

\(F=\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{2020}-\sqrt{2019}}{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2020}-\sqrt{2019}\)

\(=\sqrt{2020}-1\)

e/

\(G=\sqrt{2+\sqrt{2+\sqrt{2+...}}}\) (G>0)

\(\Rightarrow G^2=2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}\)

\(\Rightarrow G^2=2+G\)

\(\Rightarrow G^2-G-2=0\Rightarrow\left(G+1\right)\left(G-2\right)=0\)

\(\Rightarrow G=2\)

Ta có \(A=\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\)

\(\Rightarrow A^3=364+3.\sqrt[3]{182+\sqrt{33125}}.\sqrt[3]{182-\sqrt{33125}}.A\)

\(\Leftrightarrow A^3=364-3A\)

\(\Leftrightarrow\left(A-7\right)\left(A^2+7A+52\right)=0\)

Vì \(A^2+7A+52=\left(A^2+7A+\frac{49}{4}\right)+\frac{159}{4}=\left(A+\frac{7}{2}\right)^2+\frac{159}{4}>0\)

Do đó A - 7 = 0 => A = 7

\(E^3=182+\sqrt{33125}+182-\sqrt{33125}+3\sqrt[3]{182^2-33125}\left(E\right)\)

   =\(364-3E\)

\(\Rightarrow E^3+3E-364=0\) 

\(\Leftrightarrow E^3-7E^2+7E^2-49E+52E-364=0\)

\(\Leftrightarrow\left(E-7\right)\left(E^2+7E+52\right)=0\)

\(\Rightarrow E=7\)

ta có \(E^3=\left(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\right)^3\)

\(E^3=\left(182+\sqrt{33125}\right)+\left(182-\sqrt{33125}\right)+3\cdot E\cdot\sqrt[3]{33124-33125}\)

\(E^3=364-3E\)

giải phương trình \(E^3+3E-364=0\)

suy ra E= 7

Áp dụng hằng đẳng thức (a + b)³ = a³ + b³ + 3ab(a + b)

\(A^3=182+\sqrt{33125}+182-\sqrt{33125}+3\sqrt[3]{182+\sqrt{33125}}.\sqrt[3]{182-\sqrt{33125}}.\left(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\right)\)

\(A^3=364+3\sqrt[3]{182^2-33125}.A\)

A³ = 364 - 3A

<=> A³ + 3A - 364 = 0 

<=> A³ - 7A² + 7A² - 49A + 52A  364 = 0

<=> A². (A - 7) + 7A.(A - 7) + 52.(A - 7)= 0

<=> (A - 7).(A² + 7A + 52) = 0 

<=> A = 7 hoặc A² + 7A + 52 = 0  (*)

Giải (*) <=> (A+ 3,5)² + 39,75 = 0 (vô nghiệm)

Vậy A = 7

Bấm máy tinh ta được \(A=7\)nên sẽ dự đoán như sau (lưu ý \(\sqrt{33125}=25\sqrt{53}\)):

\(\hept{\begin{cases}\sqrt[3]{182-25\sqrt{53}}=\frac{7}{2}-a\sqrt{53}\\\sqrt[3]{182+25\sqrt{53}}=\frac{7}{2}+a\sqrt{53}\end{cases}}\)

Khi đó cộng lại sẽ được 7

Tìm a thì quá đơn giản: \(a=\frac{\sqrt[3]{182+25\sqrt{53}}-\frac{7}{2}}{\sqrt{53}}\)

Bấm máy tính, ta được ngay \(a=\frac{1}{2}\)

Vậy \(\sqrt[3]{182\pm25\sqrt{53}}=\frac{7}{2}\pm\frac{\sqrt{53}}{2}\)

Muốn chứng minh thì lập phương 2 vế là được.