Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{\sqrt{2}\sqrt{8-\sqrt{15}}}{\sqrt{2}\left(\sqrt{15}.\sqrt{2}-\sqrt{2}\right)}=\frac{\sqrt{16-2\sqrt{15}}}{\sqrt{2}.\sqrt{2}\left(\sqrt{15}-1\right)}\)
\(=\frac{\sqrt{15-2\sqrt{15}+1}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\frac{1}{2}\)
A=\(\sqrt{5-2\sqrt{3}.\sqrt{5}+3}-\sqrt{5+2\sqrt{5}.\sqrt{3}+3}\)
A=\(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
A=\(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
A=\(-2\sqrt{3}\)
\(A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(A=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(A=\left|\sqrt{5}-\sqrt{3}\right|-\sqrt{5}-\sqrt{3}\)
\(A=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
\(A=-2\sqrt{3}\)
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
=> \(\sqrt{4}=2>\sqrt{3}\)
=> \(A>\sqrt{3}\)
\(D=\sqrt{\frac{8+\sqrt{15}}{2}}+\sqrt{\frac{8-\sqrt{15}}{2}}\)
\(\Rightarrow D^2=\frac{8+\sqrt{15}}{2}+\frac{8-\sqrt{15}}{2}+2.\sqrt{\frac{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}{2.2}}\)
\(=8+2.\sqrt{\frac{64-15}{4}}\)
\(=8+2.\frac{7}{2}=8+7=15\)
\(\Rightarrow D=\sqrt{15}\text{ Hoặc }D=-\sqrt{15}\)
\(\text{Mà }D>0\text{ nên }D=\sqrt{15}\)
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
\(A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{3-2\sqrt{3.5}+5}-\sqrt{3+2\sqrt{3.5}+5}\)
\(=\sqrt{\left(3-5\right)^2}-\sqrt{\left(3+5\right)^2}\)
\(=|3-5|-|3+5|\)
\(=-3+5-3-5\)
\(=-6 \)
\(A=\sqrt{4+\sqrt{7}}-\sqrt{4+\sqrt{7}}\Leftrightarrow\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(\Leftrightarrow\sqrt{2}A=\sqrt{\sqrt{7}^2+2\sqrt{7}+1}-\sqrt{\sqrt{7}^2+2\sqrt{7}+1}\)
\(\Leftrightarrow\sqrt{2}A=\sqrt{7}+1-\sqrt{7}-1=0\)
\(\Leftrightarrow A=0\)
\( A = \sqrt {\dfrac{{8 + \sqrt {15} }}{2}} + \sqrt {\dfrac{{8 - \sqrt {15} }}{2}} \\ \Rightarrow {A^2} = \dfrac{{8 + \sqrt {15} }}{2} + 2\sqrt {\dfrac{{8 + \sqrt {15} }}{2}.\dfrac{{8 - \sqrt {15} }}{2}} + \dfrac{{8 - \sqrt {15} }}{2}\\ = 8 + 2\sqrt {\dfrac{{\left( {8 + \sqrt {15} } \right)\left( {8 - \sqrt {15} } \right)}}{4}} = 8 + 7 = 15 \Rightarrow A = \sqrt {15} \)
Ta có: \(A^2=\frac{8+\sqrt{15}}{2}+\frac{8-\sqrt{15}}{2}+2\sqrt{\frac{8+\sqrt{15}}{2}.\frac{8-\sqrt{15}}{2}}\)
\(A^2=8+2\sqrt{\frac{64-15}{4}}\)
\(A^2=8+2\sqrt{\frac{49}{4}}\)
\(A^2=8+7=15\)
Mà A > 0 nên \(A=\sqrt{15}\)