tình GTBT \(P=\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt: \(A=\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}>0\)
<=> \(A.\sqrt{4+\sqrt{13}}=\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}\)
<=> \(A^2\left(4+\sqrt{13}\right)=4+\sqrt{3}+4-\sqrt{3}+2\sqrt{13}\)
<=> \(A^2\left(4+\sqrt{13}\right)=2\left(4+\sqrt{13}\right)\)
<=> \(A=\sqrt{2}\)
Vậy: \(\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)
\(=\sqrt{2}+\sqrt{25-2.5.\sqrt{2}+2}\)
\(=\sqrt{2}+\left(5-\sqrt{2}\right)=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Em tham khảo đề bài và bài làm tại link: Câu hỏi của Trân Vũ Mai Ngọc - Toán lớp 9 - Học toán với OnlineMath
![](https://rs.olm.vn/images/avt/0.png?1311)
B = \(\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{8+2\sqrt{7}}{6+\sqrt{8+2\sqrt{7}}}+\frac{8-2\sqrt{7}}{6-\sqrt{8-2\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{6+\sqrt{7}+1}+\frac{\left(\sqrt{7}-1\right)^2}{6-\sqrt{7}+1}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{\sqrt{7}\left(\sqrt{7}+1\right)}+\frac{\left(\sqrt{7}-1\right)^2}{\sqrt{7}\left(\sqrt{7}-1\right)}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\sqrt{7}+1}{\sqrt{7}}+\frac{\sqrt{7}-1}{\sqrt{7}}=\frac{2\sqrt{7}}{\sqrt{7}}=2\)
=> B = \(\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{6+\sqrt{8+2\sqrt{7}}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{6-\sqrt{8-2\sqrt{7}}}=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{6+\sqrt{\left(\sqrt{7}+1\right)^2}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{6-\sqrt{\left(\sqrt{7}-1\right)^2}}\)
\(=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{7+\sqrt{7}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{7-\sqrt{7}}=\frac{\sqrt{2}\left(4+\sqrt{7}\right)\left(7-\sqrt{7}\right)}{35}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)\left(7+\sqrt{7}\right)}{35}\)
\(=\frac{21\sqrt{2}+3\sqrt{14}}{35}+\frac{21\sqrt{2}-3\sqrt{14}}{35}=\frac{42\sqrt{2}}{35}=\frac{6\sqrt{2}}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+2\sqrt{5}}{3\sqrt{5}-1}\)
=1
Câu hỏi của Trân Vũ Mai Ngọc - Toán lớp 9 - Học toán với OnlineMath