Trả lời:

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)

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

\(A=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

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

\(A=\frac{\sqrt{2}.\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=\frac{\sqrt{2}.\sqrt{4+2\sqrt{3}}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=\frac{\sqrt{2}.\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=\frac{\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=\frac{\sqrt{2}.\left(\sqrt{3}+1\right)}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=1\)

Câu trên đề sai

\(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}=\sqrt{2}\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{6}+\sqrt{2}}=1\)

Vậy nó là số nguyên

Lớn hơn hoặc bằng đấy

Giả sử a = b = 2 thì VT = 4 < VP = 16

Nhiêu đây là thấy đề sai rồi

a) \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=\sqrt[4]{25+2\sqrt{600}+24}+\sqrt[4]{25-2\sqrt{600}+24}\\ =\sqrt[4]{\left(\sqrt{25}+\sqrt{24}\right)^2}+\sqrt[4]{\left(\sqrt{25}-\sqrt{24}\right)^2}=\sqrt{\sqrt{25}+\sqrt{24}}+\sqrt{\sqrt{25}-\sqrt{24}}\\ =\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =2\sqrt{3}\)

a) \(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)\(\Leftrightarrow A=\left[\left(\sqrt{57}+6\right)+\left(3\sqrt{6}+\sqrt{38}\right)\right]\left[\left(\sqrt{57}+6\right)-\left(3\sqrt{6}+\sqrt{38}\right)\right]\)\(\Leftrightarrow A=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)

\(\Leftrightarrow A=57+12\sqrt{57}+36-54-12\sqrt{57}-38\)

\(\Leftrightarrow A=1\)

b) \(B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+\left(2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)

\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

\(\Leftrightarrow B=\dfrac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)

\(\Leftrightarrow B=\dfrac{\sqrt{8+4\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)

\(\Leftrightarrow B=\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{\sqrt{6}+\sqrt{2}}\)

\(\Leftrightarrow B=\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=1\)

c)\(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{3^2-2\times3\times2\sqrt{5}+\left(2\sqrt{5}\right)^2}}}\)

\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)

3 bài đầu dễ tự làm nhé.

Bài 4:

\(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)

\(=\dfrac{\sqrt{\left(1-\sqrt{2}\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(1+\sqrt{2}\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)

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

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

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

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

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

\(=0+2\)

\(=2\)

Vậy B là số tự nhiên.

1.

a) nhân cả tử lẫn mẫu với 1+ \(\sqrt{2}-\sqrt{5}\)

b) tương tự a

2.

a) tách 29 = 20 + 9 là ra hằng đẳng thức, tiếp tục.

1.

a) \(\dfrac{1}{1+\sqrt{2}+\sqrt{5}}=\dfrac{1+\sqrt{2}-\sqrt{5}}{\left(1+\sqrt{2}+\sqrt{5}\right)\left(1+\sqrt{2}-\sqrt{5}\right)}\)

=\(\dfrac{1+\sqrt{2}-\sqrt{5}}{\left(1+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}=\dfrac{1+\sqrt{2}-\sqrt{5}}{1+2\sqrt{2}+2-5}\)

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

b) \(\dfrac{1}{\sqrt{x}+\sqrt{x+1}}=\dfrac{\sqrt{x}-\sqrt{x+1}}{\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x}-\sqrt{x+1}\right)}\)

=\(\dfrac{\sqrt{x}-\sqrt{x+1}}{\left(\sqrt{x}\right)^2-\left(\sqrt{x+1}\right)^2}=\dfrac{\sqrt{x}-\sqrt{x+1}}{x-x-1}=\dfrac{\sqrt{x}-\sqrt{x+1}}{-1}=-\sqrt{x}+\sqrt{x+1}\)

2.

a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)

=\(\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)=\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

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

b)\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)

=\(\sqrt{6+2\sqrt{4-\sqrt{12}}}\)

=\(\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

=\(\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}\)

=\(\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

c) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

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

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

làm giống câu a

3. a=\(\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

=\(\sqrt{3-\sqrt{5}}\left(3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\)

=\(\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)

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

=\(\sqrt{6-2\sqrt{5}}\left(2\sqrt{5}+2\right)=\left(\sqrt{5}-1\right)\left(2\sqrt{5}+2\right)\)

=\(10-2\sqrt{5}+2\sqrt{5}-2=8\)

vậy a là số tự nhiên

a) \(5\sqrt{27}+3\sqrt{48}-2\sqrt{12}-6\sqrt{3}\)

\(=15\sqrt{3}+12\sqrt{3}-4\sqrt{3}-6\sqrt{3}\)

\(=17\sqrt{3}\)

b) \(\dfrac{2}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+6\sqrt{3}\)

\(=6-3\sqrt{3}+4+\sqrt{3}+6\sqrt{3}\)

\(=4\sqrt{3}+10\)

Bài 2

\(P=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{4-\sqrt{12}}}}{\sqrt{6}-\sqrt{2}}\)

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

\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

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

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

\(=\frac{\sqrt{2}\cdot\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)

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

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

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

Vậy P là một số nguyên