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b: \(=\dfrac{\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{\sqrt{2}}=\dfrac{3+\sqrt{5}+3-\sqrt{5}}{2}=\dfrac{6}{\sqrt{2}}=3\sqrt{2}\)
c: \(=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
\(\frac{1}{\sqrt{2k+1+2\sqrt{k^2+k}}}=\frac{1}{\sqrt{k+1+2\sqrt{k\left(k+1\right)}+k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}\)
\(=\frac{\sqrt{k+1}-\sqrt{k}}{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{k+1-k}=\sqrt{k+1}-\sqrt{k}\)
Do đó ta có:
\(A=\frac{1}{\sqrt{3+2\sqrt{2}}}+...+\frac{1}{\sqrt{2n+1+2\sqrt{n^2+n}}}\)
\(A=\sqrt{2}-\sqrt{1}+...+\sqrt{n+1}-\sqrt{n}\)
\(A=\sqrt{n+1}-1\)
Với \(n=2018\)ta có: \(A=\sqrt{2019}-1\).
1: \(=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
2: \(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
a, \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
b, \(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{4-2.2\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+1+2-\sqrt{2}=3\)
câu 1 đã làm
câu 2
\(\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(\Leftrightarrow\sqrt{2}+1+\sqrt{2}-2\Leftrightarrow2\sqrt{2}-1\)
Ta đặt: \(A=\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\right)^2\)
<=> \(A^2=\sqrt{7}-\sqrt{3}-2\sqrt{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\sqrt{7}+\sqrt{3}\)
<=> \(A^2=2\sqrt{7}-2\sqrt{7-3}\)
<=> \(A^2=2\sqrt{7}-2\sqrt{4}=2\left(\sqrt{7}-2\right)\)
=> \(A=\sqrt{2\left(\sqrt{7}-2\right)}\)
Thay vào ta được:
\(\frac{\sqrt{2\left(\sqrt{7}-2\right)}}{\sqrt{\sqrt{7}-2}}=\sqrt{2}\)