\(\Rightarrow A+B< 3+5=8\)

mặt khác ta có A+B>\(\sqrt{20}+\sqrt[3]{24}=7.3566....>7\)\(\Rightarrow\left[A+b\right]=7\)

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

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

\(=4-3\cdot A\)

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

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

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

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

\(\Leftrightarrow A=1\)

Ta có:
\(A< \sqrt{20+\sqrt{20+...+\sqrt{20+\sqrt{25}}}}\)
\(\Leftrightarrow A< \sqrt{25}=5\)(1)
\(B< \sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24+\sqrt[3]{27}}}}\)
\(\Leftrightarrow B< \sqrt[3]{27}=3\)(2)
Từ (1) và (2) suy ra A+B<5+3=8
Ta có:
\(A>\sqrt{19,36}=4,4\)(3)
\(B>\sqrt[3]{17,576}=2,6\)(4)
Từ (3) và (4) suy ra A+B>4,4+2,6=7
Vậy 7<A+B<8

a)\(A=^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\)

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

\(=20+14\sqrt{2}+20-14\sqrt{2}\)

\(+3\left(\text{​​}^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\right)\left(^3\sqrt{20+14\sqrt{2}}.^3\sqrt{20-14\sqrt{2}}\right)\)

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

\(\Rightarrow A^3=40+3.A.2\)

=> \(A^3-6A-40=0\)

<=> \(A^3-16A+10A-40=0\)

<=> \(A\left(A-4\right)\left(A+4\right)+10\left(A-4\right)=0\)

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

<=> A = 4 ( vì \(A^2+4A+10=\left(A+2\right)^2+6>0\))

Vậy A = 4.

b/ \(B=^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\)

=> \(B^3=\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right)^3\)

\(=26+15\sqrt{3}-26+15\sqrt{3}\)

\(-3\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right).^3\sqrt{26+15\sqrt{3}}.^3\sqrt{26-15\sqrt{3}}\)

\(=30\sqrt{3}-3B.1\)

=> \(B^3+3B-30\sqrt{3}=0\)

<=> \(B^3-12B+15B-30\sqrt{3}=0\)

<=> \(B\left(B-2\sqrt{3}\right)\left(B+2\sqrt{3}\right)+15\left(B-2\sqrt{3}\right)=0\)

<=> \(\left(B-2\sqrt{3}\right)\left(B^2+2\sqrt{3}B+15\right)=0\)

<=> \(B-2\sqrt{3}=0\)( vì \(B^2+2\sqrt{3}B+15=\left(B+\sqrt{3}\right)^2+12>0\))

<=> \(B=2\sqrt{3}\)

B = \(\sqrt{\sqrt{75-2.2\sqrt{2}.5\sqrt{3}+8}+\sqrt{50-2.2\sqrt{3}.5\sqrt{2}+12}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)

   = \(\sqrt{\sqrt{\left(5\sqrt{3}-2\sqrt{2}\right)^2}+\sqrt{\left(5\sqrt{2}+2\sqrt{3}\right)^2}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)

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

   = \(\sqrt{3\sqrt{3}+3\sqrt{2}}.\sqrt{3\sqrt{3}-3\sqrt{2}}=\sqrt{\left(3\sqrt{3}+3\sqrt{2}\right)\left(3\sqrt{3}-3\sqrt{2}\right)}\)

   = \(\sqrt{27-18}=\sqrt{9}=3\)

\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)

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

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

\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)

\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)

\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)

\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)

\(\Rightarrow X^3=2-X\)

\(\Rightarrow X^3+X-2=0\)

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

\(\Rightarrow X=1\) (do \(X^2+2X+2=\left(X+1\right)^2+1>0\) \(\forall X\))