1. Trục căn thức ở mẫu:
a) \(\dfrac{1}{1+\sqrt{2}+\sqrt{5}} \)
b) \(\dfrac{1}{\sqrt{x}+\sqrt{x+1}}\)
2. Tính:
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
b) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
c) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
3. Cho a = \(\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
Chứng minh rằng a là số tự nhiên.
4. Cho b = \(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
b có phải là số tự nhiên không?
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
