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50) \(\sqrt{98-16\sqrt{3}}=4\sqrt{6}-\sqrt{2}\)
51) \(\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
52) \(\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
53) \(\sqrt{5-\sqrt{21}}=\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)
54) \(\sqrt{6-\sqrt{35}}=\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{2}\)
55) \(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
56) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
57.\(\sqrt{8-\sqrt{55}}=\sqrt{\dfrac{16-2.\sqrt{5}.\sqrt{11}}{2}}=\sqrt{\dfrac{\sqrt{11}^2-2.\sqrt{5}.\sqrt{11}+\left(\sqrt{5}\right)^2}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{11}-\sqrt{5}\right)^2}{2}}=\dfrac{\left|\sqrt{11}-\sqrt{5}\right|}{\sqrt{2}}=\dfrac{\sqrt{11}-\sqrt{5}}{\sqrt{2}}\)
58. \(\sqrt{7+\sqrt{33}}=\sqrt{\dfrac{14+2\sqrt{3}.\sqrt{11}}{2}}=\sqrt{\dfrac{\left(\sqrt{11}\right)^2+2\sqrt{3}.\sqrt{11}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{11}+\sqrt{3}\right)^2}{2}}=\dfrac{\left|\sqrt{11}+\sqrt{3}\right|}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{3}}{\sqrt{2}}\)
mấy câu dưới bạn cũng làm tương tự thôi
60) \(\sqrt{7-3\sqrt{5}}=\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}=\dfrac{3\sqrt{2}-\sqrt{10}}{2}\)
61) \(\sqrt{23+3\sqrt{5}}=\dfrac{\sqrt{46+6\sqrt{5}}}{\sqrt{2}}=\dfrac{3\sqrt{5}+1}{\sqrt{2}}=\dfrac{3\sqrt{10}+\sqrt{2}}{2}\)
62) \(\sqrt{7-\sqrt{33}}=\dfrac{\sqrt{14-2\sqrt{33}}}{\sqrt{2}}=\dfrac{\sqrt{11}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{22}-\sqrt{6}}{2}\)
63) \(\sqrt{8+\sqrt{55}}=\dfrac{\sqrt{16+2\sqrt{55}}}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{22}+\sqrt{10}}{2}\)
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) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
a: \(=9\sqrt{2}-4\sqrt{2}+4\sqrt{2}+9\sqrt{2}=18\sqrt{2}\)
b: \(=8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)
c: \(=2\sqrt{21}\)
Bài 1:
1) Ta có: \(3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
2) Ta có: \(8+2\sqrt{7}\)
\(=7+2\cdot\sqrt{7}\cdot1+1\)
\(=\left(\sqrt{7}+1\right)^2\)
3) Ta có: \(x-2\sqrt{x-1}\)
\(=x-1-2\cdot\sqrt{x-1}\cdot1+1\)
\(=\left(\sqrt{x-1}-1\right)^2\)
4) Ta có: \(6-4\sqrt{2}\)
\(=4-2\cdot2\cdot\sqrt{2}+2\)
\(=\left(2-\sqrt{2}\right)^2\)
5) Ta có: \(7+4\sqrt{3}\)
\(=4+2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2+\sqrt{3}\right)^2\)
6) Ta có: \(9-4\sqrt{5}\)
\(=5-2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}-2\right)^2\)
7) Ta có: \(10+2\sqrt{21}\)
\(=7+2\cdot\sqrt{7}\cdot\sqrt{3}+3\)
\(=\left(\sqrt{7}+\sqrt{3}\right)^2\)
8) Ta có: \(49+20\sqrt{6}\)
\(=25+2\cdot5\cdot2\sqrt{6}+24\)
\(=\left(5+2\sqrt{6}\right)^2\)
60) \(\sqrt{7-3\sqrt{5}}=\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{2}}=\dfrac{\left(3-\sqrt{5}\right)}{\sqrt{2}}=\dfrac{3\sqrt{2}-\sqrt{10}}{2}\)
59) \(\sqrt{6+\sqrt{35}}=\dfrac{\sqrt{12+2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{14}+\sqrt{10}}{2}\)
61) \(\sqrt{23+3\sqrt{5}}=\dfrac{\sqrt{46+6\sqrt{5}}}{\sqrt{2}}=\dfrac{3\sqrt{5}+1}{\sqrt{2}}=\dfrac{3\sqrt{10}+\sqrt{2}}{2}\)
62) \(\sqrt{7-\sqrt{33}}=\dfrac{\sqrt{14-2\sqrt{33}}}{\sqrt{2}}=\dfrac{\sqrt{11}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{22}-\sqrt{6}}{2}\)
63) \(\sqrt{8+\sqrt{55}}=\dfrac{\sqrt{16+2\sqrt{55}}}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{22}+\sqrt{10}}{2}\)
Câu 64 và 38 bạn làm đc ko ạ? Vs cả bạn giải chi tiết hơn giùm mình đc ko ạ?