Trả lời:

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

\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)

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

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

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

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

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

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

\(A=\sqrt{1}\)

\(A=1\)

\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(3+2\sqrt{6}+2\right).\left(49-20\sqrt{6}\right).\sqrt{3-2\sqrt{6}+2}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{33}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right).\left(49-20\sqrt{6}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(3-2\right).\left(49\sqrt{3}-60\sqrt{2}+49\sqrt{2}-40\sqrt{3}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{1.\left(9\sqrt{3}-11\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=1\)

a) Ta có: \(\sqrt{29-12\sqrt{5}}=\sqrt{20-12\sqrt{5}+9}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)

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

\(\Rightarrow\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{3-\left(2\sqrt{5}-3\right)}=\sqrt{3-2\sqrt{5}+3}\)

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

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

\(\Leftrightarrow A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

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

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

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+9}}}\)

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

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

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

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

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

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

\(=\sqrt{1}\)

\(=1\)

Vậy A là số tự nhiên

nhưng mà olm chọn rồi thì chọn nhiều đến mấy cũng cộng dc 3 điểm

trơi , 1 bài mà triệu đang với ngọc vĩ , chat với nhau hết

\(Q=\sqrt{\sqrt{5}-1}\left(\sqrt{8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}}-\sqrt{7-\sqrt{20}}\right)\)

\(\Rightarrow\)\(Q^2=\left(\sqrt{5}-1\right)\left(8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}+7-\sqrt{20}-2\sqrt{\left(7-\sqrt{20}\right)\left(8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}\right)\left(8-\sqrt{5}\right)+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{66-23\sqrt{5}+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(49-28\sqrt{5}+20\right)+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}+\left(5\sqrt{5}-3\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}\right)^2+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}+\left(5\sqrt{5}-3\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}+\sqrt{5\sqrt{5}-3}\right)^2}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\left(7-2\sqrt{5}+\sqrt{5\sqrt{5}-3}\right)\right)\)

\(=\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)\)\(=4\)

\(\Rightarrow Q^2=4\) \(\Rightarrow Q\) nguyên 

\(\sqrt{29}\)

a) A=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

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

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

b) B=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

=\(\left(4\sqrt{10}+\sqrt{150}-4\sqrt{6}-\sqrt{90}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

=\(\left(4\sqrt{10}+5\sqrt{6}-4\sqrt{6}-3\sqrt{10}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

=\(\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

=\(5-\sqrt{15}+\sqrt{15}-3=2\)(đpcm)

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,

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=a-2\sqrt{ab}+b=\left(\sqrt{a}-\sqrt{b}\right)^2\)

b,

A=\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+2\sqrt{12}}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-1-\sqrt{12}}}}{\sqrt{6}+\sqrt{2}}\)\(=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{2}\sqrt{4+2\sqrt{3}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=1\)

B=

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

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

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

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

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

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

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

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