Chứng minh rằng biểu thức sau nhận giá trị nguyên:
\(B=\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3-11\sqrt{2}}}\)
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Chứng minh rằng biểu thức sau nhận giá trị nguyên:
\(B=\frac{\left(5+2.\sqrt{6}\right)\left(49-20.\sqrt{6}\right)\sqrt{5-2.\sqrt{6}}}{9.\sqrt{3-11.\sqrt{2}}}\)
Chứng minh rằng các số sau đây là số nguyên:
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
B = \(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
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 )
có ai biết giải bài này không giúp mình với mình đang cần gấp, xin cảm ơn
Bài 20: rút gọn
1, \(\sqrt{9-4\sqrt{5}}.\sqrt{9+4\sqrt{5}}\)
2, \(\left(2\sqrt{2}-6\right).\sqrt{11+6\sqrt{2}}\)
3, \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)\)
4, \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
5, \(\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}\)
Bài 21: rút gọn
1, \(5\sqrt{\dfrac{1}{5}}\) 2, \(\dfrac{12}{5}\sqrt{\dfrac{5}{4}}\)
3, \(\dfrac{30}{5\sqrt{6}}\) 4, \(\dfrac{20}{2\sqrt{5}}\)
5, \(\dfrac{2-\sqrt{2}}{\sqrt{2}}\)
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
Chứng minh rằng:
a)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\) là số nguyên
b)\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Tính giá trị các biểu thức sau:
a) \(A=\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
b) \(A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
c) \(A=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(A=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2}{9\sqrt{3}-11\sqrt{2}}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(9\sqrt{3}+11\sqrt{3}\right)\left(5-2\sqrt{6}\right)^2\)
\(=\left(49+20\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2=\left(5+2\sqrt{6}\right)^2\left(5-2\sqrt{6}\right)^2=1\)
\(A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{4+5}=3\)
\(A=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\)
Giúp mình bài này với mình đang cần gấp
Rút gọn các biểu thức sau:
C=\(\frac{\sqrt{6+2\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}\)\(-\frac{\sqrt{6-2\left(\sqrt{6}-\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}\)
D=\(\frac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}\)
Rút gọn biểu thức
A=\(\sqrt{6+3\sqrt{3}}\)+\(\sqrt{6-3\sqrt{3}}\)
B=\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Bài 1: Tính giá trị của biểu thức:\(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 2: Chứng minh rằng các biểu thức sau có giá trị là số nguyên
A = \(\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
B = \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
giải giúp mình, cần gấp
\(A=\frac{\sqrt{3}-3}{\sqrt{2-\sqrt{3}}+2\sqrt{2}}+\frac{\sqrt{3}+3}{\sqrt{2+\sqrt{3}}-2\sqrt{2}}\)
B= \(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)