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17 tháng 12 2023
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
N
25 tháng 10 2021
\(\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{11\left(4-\sqrt{5}\right)}{16-5}=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)
MV
3 tháng 7 2018
a.
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\\ =\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\\ =\sqrt{81-17}\\ =\sqrt{64}\\=8\)
3 tháng 7 2018
\(a.VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=8=VP\)
\(b.\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=3\sqrt{3}-\sqrt{2}\) ( thiếu đề )
\(VT=\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=\dfrac{1}{3-2\sqrt{3}.\sqrt{2}+2}+\dfrac{2}{3+2\sqrt{3}.\sqrt{2}+2}=\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}+2\sqrt{3}-2\sqrt{2}=3\sqrt{3}-\sqrt{2}=VP\)
AH
Akai Haruma
Giáo viên
12 tháng 6 2021
Lời giải:
Gọi tổng trên là $A$. Ta có:
\(2A>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)
\(2A>\frac{\sqrt{2}-1}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3})}+...+\frac{\sqrt{81}-\sqrt{80}}{(\sqrt{80}+\sqrt{81})(\sqrt{81}-\sqrt{80})}\)
\(2A>(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{81}-\sqrt{80})\)
\(2A>\sqrt{81}-1=8\Rightarrow A>4\)
Ta có đpcm.
12 tháng 7 2018
\(1.\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\sqrt{17-4\sqrt{5+2.2\sqrt{5}+4}}=\sqrt{17-4\left(\sqrt{5}+2\right)}=\sqrt{5-2.2\sqrt{5}+4}=\sqrt{5}-2\)
\(2.\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{17-6\sqrt{2+2\sqrt{2}+1}}=\sqrt{17-6\left(\sqrt{2}+1\right)}=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)\(3.\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}=\sqrt{3+\sqrt{3-2\sqrt{3}+1}}=\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(4.\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}=\sqrt{25+2.5\sqrt{2}+2}.\left(5-\sqrt{2}\right)=\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)=5-2=3\)
\(\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}=\dfrac{2-\sqrt{5}}{\sqrt{9-4\sqrt{5}}}=\dfrac{2-\sqrt{5}}{\sqrt{5}-2}=-1\)
\(\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{9+4\sqrt{5}}}}=\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\\ =\dfrac{2-\sqrt{5}}{\sqrt{17-4\left(\sqrt{5}+2\right)}}=\dfrac{2-\sqrt{5}}{\sqrt{9-4\sqrt{5}}}\\ =\dfrac{2-\sqrt{5}}{\sqrt{\left(\sqrt{5}-2\right)^2}}=\dfrac{2-\sqrt{5}}{\sqrt{5}-2}=-1\)