\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}-1-\sqrt{5}-1=-2\)
Vậy \(A\in Z\)
Làm tương tự với B.
CM A thuoc Z va B thuoc Z voi :
A = \(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
B = \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
Chững minh B\(\in\)Z với:
B=\(\frac{3-2\sqrt{2}}{\sqrt{17-12\sqrt{2}}}\)\(-\)\(\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
1/ \(\frac{2}{3-\sqrt{7}}\sqrt{\frac{6\sqrt{2}-2\sqrt{14}}{3\sqrt{2}+\sqrt{14}}}\)
2/ \(\sqrt{6+2\sqrt{\sqrt{5}-\sqrt{13-\sqrt{48}}}}\)
3/ \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
4/ \(\frac{24}{\sqrt{7}+1}+\frac{4}{3+\sqrt{7}}-\frac{3}{\sqrt{7}+2}\left(4-\sqrt{7}\right)\)
5/ \(\sqrt{7-3\sqrt{5}}\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
Tính giá trị của biểu thức:
a)A=\(\sqrt{\left(2-\sqrt{5}\right)^2}\) +\(\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
b)B=\(\sqrt{6+2\sqrt{5}}\) - \(\sqrt{6-2\sqrt{5}}\)
c)C=\(\sqrt{17+12\sqrt{2}}\) + \(\sqrt{17-12\sqrt{2}}\)
Phương pháp 5. Biến đổi về dạng tổng các bình phương \(A^2+B^2+C^2=0\)
a \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)
b \(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)
c \(9x+17=6\sqrt{8x+1}+4\sqrt{x+3}\)
d \(\sqrt{x}+2\sqrt{x+3}=x+4\)
e\(\sqrt{3-x}+2\sqrt{3x-2}-3=x\)
Tính P=\(a^3+b^3-3\left(a+b\right)+2012\)
Biết \(a=\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}};b=\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\)
Bài 1: Chứng minh:
a) \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}=0\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
c) \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
d) \(\sqrt{6+\sqrt{6+\sqrt{6+,,,+\sqrt{6}}}}=3\)
a : \(\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{17+12\sqrt{2}}\)
b : \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
c : \(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\)
tính P= a3 +b3- 3(a+b)+2018. Biết
a=\(\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}}\)
\(b=\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\)
