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\(\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{\sqrt{2}\sqrt{8-\sqrt{15}}}{\sqrt{2}\left(\sqrt{15}.\sqrt{2}-\sqrt{2}\right)}=\frac{\sqrt{16-2\sqrt{15}}}{\sqrt{2}.\sqrt{2}\left(\sqrt{15}-1\right)}\)
\(=\frac{\sqrt{15-2\sqrt{15}+1}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\sqrt{\frac{8+\sqrt{15}}{2}}+\sqrt{\frac{8-\sqrt{15}}{2}}\)
\(\Rightarrow D^2=\frac{8+\sqrt{15}}{2}+\frac{8-\sqrt{15}}{2}+2.\sqrt{\frac{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}{2.2}}\)
\(=8+2.\sqrt{\frac{64-15}{4}}\)
\(=8+2.\frac{7}{2}=8+7=15\)
\(\Rightarrow D=\sqrt{15}\text{ Hoặc }D=-\sqrt{15}\)
\(\text{Mà }D>0\text{ nên }D=\sqrt{15}\)
D=√8+√152 +√8−√152
⇒D2=8+√152 +8−√152 +2.√(8+√15)(8−√15)2.2
=8+2.√64−154
=8+2.72 =8+7=15
⇒D=√15 Hoặc D=−√15
Mà D>0 nên D=√15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7}+2\sqrt{10}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{7}+2\sqrt{10}}\)(tách ra hằng đẳng thức)
\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{7}+2\sqrt{10}}\)
\(=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{7}+2\sqrt{10}}\)
p/s: tách đến đây là em bí rồi ạ :(
ko sao a~, ra là tớ chép đề bài sai, haizzz
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(A^2=2+2.\frac{\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}}{2}\)
\(2+\sqrt{64-15}=2+\sqrt{49}=2+7=9\) mà A>0
=> A=3
2) \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\)
\(A=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)
\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)
\(A^2=\left(4+\sqrt{15}\right)\left(16-4\sqrt{15}\right)\)
\(=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\)
Mà A >0
=> A=2
Mà 4>3
=> \(\sqrt{4}=2>\sqrt{3}\)
=> \(A>\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4: \(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\)
\(=2\sqrt{3}\)
4) \(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{5}+\sqrt{3}-\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}\)
5) \(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{2}+\sqrt{3}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\)
\(=\frac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}\)
\(=\frac{12\sqrt{2}-2\sqrt{18}}{6}=\frac{6\sqrt{2}}{6}=\sqrt{2}\)
\(\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{1}{2}\) NHA
Nguyễn Thị My Na !