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NP
2
15 tháng 6 2019
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}+8-2\sqrt{10+2\sqrt{5}}+}2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(A^2=16+2\left[64-4\left(10+2\sqrt{5}\right)\right]\)
\(A^2=16+128-8\left(10+2\sqrt{5}\right)\)
\(A^2=144-80-16\sqrt{5}\)
\(A^2=64-16\sqrt{5}\)
15 tháng 6 2019
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2.\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)
\(=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}=16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=2+2.\sqrt{2}.\sqrt{10}+10\)
\(=\left(\sqrt{2}+\sqrt{10}\right)^2\)
=> \(A=\sqrt{2}+\sqrt{10}\)
K
1
15 tháng 6 2019
Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath
Bạn tham khảo link này!
16 tháng 8 2015
Biến đổi vế trái ta có :
\(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
= \(\sqrt{2}\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)\)
Đặt A = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
A^2 = \(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
= 8 + \(2\sqrt{16-\left(10-2\sqrt{5}\right)}\)
= \(8+2\sqrt{16-10+2\sqrt{5}}\)
= \(8+2\sqrt{6+2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
=> A = \(\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
=> \(\sqrt{2}A=\sqrt{2}\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}=VP\) ( ĐPCM)
25 tháng 6 2021
\(Q=\sqrt{\sqrt{5}-1}\left(\sqrt{8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}}-\sqrt{7-\sqrt{20}}\right)\)
\(\Rightarrow\)\(Q^2=\left(\sqrt{5}-1\right)\left(8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}+7-\sqrt{20}-2\sqrt{\left(7-\sqrt{20}\right)\left(8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}\right)}\right)\)
\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}\right)\left(8-\sqrt{5}\right)+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}}\right)\)
\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{66-23\sqrt{5}+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}}\right)\)
\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(49-28\sqrt{5}+20\right)+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}+\left(5\sqrt{5}-3\right)}\right)\)
\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}\right)^2+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}+\left(5\sqrt{5}-3\right)}\right)\)
\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}+\sqrt{5\sqrt{5}-3}\right)^2}\right)\)
\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\left(7-2\sqrt{5}+\sqrt{5\sqrt{5}-3}\right)\right)\)
\(=\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)\)\(=4\)
\(\Rightarrow Q^2=4\) \(\Rightarrow Q\) nguyên
12 tháng 10 2023
a: \(\left(3+\sqrt{2}\right)^2=3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2\)
\(=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b: \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c: \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d: \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-2\cdot3\sqrt{5}\cdot2+4}-\sqrt{45+2\cdot3\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
MH
12 tháng 10 2023
a) \(\left(3+\sqrt{2}\right)^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
17 tháng 12 2023
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
12 tháng 6 2023
a: \(=4\sqrt[3]{2}-9\sqrt[3]{2}++6\sqrt[3]{2}=\sqrt[3]{2}\)
b: \(=6\sqrt[3]{3}-15\sqrt[3]{3}+16\sqrt[3]{3}=7\sqrt[3]{3}\)
c: \(=-7\sqrt[3]{3}+3\sqrt[3]{3}+6\sqrt[3]{3}=2\sqrt[3]{3}\)
d: \(=8\sqrt[3]{5}-10\sqrt[3]{5}+2=-2\sqrt[3]{5}+2\)
\(\sqrt{\sqrt{5}-\sqrt{8-\sqrt{81-8\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{8-\left(4\sqrt{5}-1\right)}}\)\(=\sqrt{\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-2\right)}=\sqrt{2}\)