rút gọn M= 1.3.5.9....29/16.17.18....30
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Ta có: \(\sqrt{29+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2+2\sqrt{2}+1}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2+2\sqrt{2}\cdot1+1}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{\left(\sqrt{2}+1\right)^2}}+5\sqrt{2}\)
\(=\sqrt{29+30\left(\sqrt{2}+1\right)}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2}+30}+5\sqrt{2}\)
\(=\sqrt{9+2\cdot3\cdot5\sqrt{2}+50}+5\sqrt{2}\)
\(=\sqrt{\left(3+5\sqrt{2}\right)^2}+5\sqrt{2}\)
\(=3+5\sqrt{2}+5\sqrt{2}=3+10\sqrt{2}\)
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\(A=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}=\frac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}=5.\left(-27\right)=-135\)
Vậy \(A=-135\)
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\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3\sqrt{20}+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
\(b,=\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\) \(=\sqrt{3+30\sqrt{2+\sqrt{8+2\sqrt{8}+1}}}\)
\(=\sqrt{3+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)\(=\sqrt{3+30\sqrt{3+\sqrt{8}}}=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{3+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{3+30\sqrt{2}+30}=\sqrt{33+30\sqrt{2}}\)
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
=1
b) Ta có: \(\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{3+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{33+30\sqrt{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{43}}\)
\(A=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}\)
\(A=\frac{5^{30}.7^{48}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}=5.7^3.\left(1-7.2^2\right)=1715.\left(-27\right)=-46305\)
\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}\left(2^4\right)^2.7^{43}}=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}=\frac{7^{48}.5^{30}.2^8\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}\)
=\(7^5.5.\left(-27\right)=-2268945\)
Biểu thức của bạn không có qui luật a~ mẫu mình không nói nhưng tử khi bạn đặt dấu ... là bạn đã đặt dấu chấm hết cho bài mình rồi
như 1.3.5 theo qui tắc cách đều với khoảng cách là 2 nhưng sau đó lại x9 với khoảng cách là 4 bạn hãy xem lại đề xem có thiếu x7 không nha!
\(M=\frac{\left(1.3.5....15\right).\left(17.19...29\right)}{\left(16.18.20...30\right).\left(17.19...29\right)}=\frac{\left(1.3.5.7\right).\left(9.11.13.15\right)}{\left(16.20.24.28\right)\left(18.22.26.30\right)}\)
=> \(M=\frac{1.3.5.7}{16.20.24.28}.\frac{1}{8}=\frac{1}{16.4.8.4.8}=\frac{1}{2^4.2^2.2^3.2^2.2^3}=\frac{1}{2^{14}}\)
Đáp số: \(M=\frac{1}{2^{14}}\)