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Bài làm:
a) \(A=\sqrt{4}-2\sqrt{3}+\sqrt{7}-4\sqrt{3}\)
\(A=2+\sqrt{7}-6\sqrt{3}\)
b) \(B=\sqrt{3}+\sqrt{8}+\sqrt{3}-\sqrt{8}\)
\(B=2\sqrt{3}\)
b,\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\) \(=\sqrt{8\sqrt{3}}-2\sqrt{50\sqrt{3}}+4\sqrt{8\sqrt{3}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}\)
\(=0\)
d,\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{2}(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}})\)
\(\sqrt2A=\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
\(\sqrt2A=\sqrt{(\sqrt5-1)^2}\) \(+\sqrt{(\sqrt5+1)^2}\) \(=\sqrt5-1 +\sqrt5+1=2\sqrt5\)
\(\Rightarrow A=\dfrac{2\sqrt5}{\sqrt2}\) \(=\sqrt{10}\)
a. \(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{3\sqrt{5}-3+5-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=\frac{2\left(\sqrt{5}+1\right)}{2\left(\sqrt{5}+1\right)}=1\)
a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2\sqrt{3}-1\right|+\left|2-\sqrt{3}\right|\)
\(=2\sqrt{3}-1+2-\sqrt{3}=\sqrt{3}+1\)
b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)
c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4\)
e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)
f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)
a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))
=\(\sqrt{2006}^2-\sqrt{2005}^2\)
=2006-2005
=1
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
Ta có : \(A=\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{2^2-2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1+2-\sqrt{3}=1\)
Ta có : \(B=\sqrt{3+\sqrt{8}+\sqrt{3-\sqrt{8}}}\)
\(=\sqrt{3+\sqrt{8}+\sqrt{2-2\sqrt{2}+1}}\)\(=\sqrt{3+\sqrt{8}+\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}+\sqrt{2}-1}\) \(=\sqrt{2+3\sqrt{2}}\)