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1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
Lời giải:
\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{\frac{6-2\sqrt{5}}{2}}+\sqrt{\frac{6+2\sqrt{5}}{2}}\)
\(=\sqrt{\frac{5-2\sqrt{5}+1}{2}}+\sqrt{\frac{5+2\sqrt{5}+1}{2}}=\sqrt{\frac{(\sqrt{5}-1)^2}{2}}+\sqrt{\frac{(\sqrt{5}+1)^2}{2}}\)
\(=\frac{\sqrt{5}-1}{\sqrt{2}}+\frac{\sqrt{5}+1}{\sqrt{2}}=2.\frac{\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
\(B=\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)
\(=\sqrt{18+2\sqrt{18.3}+3}+\sqrt{18-2\sqrt{18.3}+3}\)
\(=\sqrt{(\sqrt{18}+\sqrt{3})^2}+\sqrt{(\sqrt{18}-\sqrt{3})^2}\)
\(=\sqrt{18}+\sqrt{3}+\sqrt{18}-\sqrt{3}=2\sqrt{18}=6\sqrt{2}\)
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\(C=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow C^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{(4+\sqrt{10+2\sqrt{5}})(4-\sqrt{10+2\sqrt{5}})}\)
\(8+2\sqrt{4^2-(10+2\sqrt{5})}=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{(\sqrt{5}-1)^2}\)
\(=8+2(\sqrt{5}-1)=6+2\sqrt{5}=(\sqrt{5}+1)^2\)
\(\Rightarrow C=\sqrt{5}+1\)
Lời giải:
\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{\frac{6-2\sqrt{5}}{2}}+\sqrt{\frac{6+2\sqrt{5}}{2}}\)
\(=\sqrt{\frac{5-2\sqrt{5}+1}{2}}+\sqrt{\frac{5+2\sqrt{5}+1}{2}}=\sqrt{\frac{(\sqrt{5}-1)^2}{2}}+\sqrt{\frac{(\sqrt{5}+1)^2}{2}}\)
\(=\frac{\sqrt{5}-1}{\sqrt{2}}+\frac{\sqrt{5}+1}{\sqrt{2}}=2.\frac{\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
\(B=\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)
\(=\sqrt{18+2\sqrt{18.3}+3}+\sqrt{18-2\sqrt{18.3}+3}\)
\(=\sqrt{(\sqrt{18}+\sqrt{3})^2}+\sqrt{(\sqrt{18}-\sqrt{3})^2}\)
\(=\sqrt{18}+\sqrt{3}+\sqrt{18}-\sqrt{3}=2\sqrt{18}=6\sqrt{2}\)
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\(C=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow C^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{(4+\sqrt{10+2\sqrt{5}})(4-\sqrt{10+2\sqrt{5}})}\)
\(8+2\sqrt{4^2-(10+2\sqrt{5})}=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{(\sqrt{5}-1)^2}\)
\(=8+2(\sqrt{5}-1)=6+2\sqrt{5}=(\sqrt{5}+1)^2\)
\(\Rightarrow C=\sqrt{5}+1\)
a, \(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=\left(-\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=-1\)
b.\(\sqrt{16+2\sqrt{16.5}+5}+\sqrt{16-2\sqrt{16.5}+5}=\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}=8\)
d,dat \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\Rightarrow A^2=4+\sqrt{7}+2\sqrt{16-7}+4-\sqrt{7}\)\(A^2=8+6=14\Rightarrow A=\sqrt{14}\)
C,\(\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}=\sqrt{17-4\left(2+\sqrt{5}\right)}=\sqrt{17-8-4\sqrt{5}}=\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
