Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}+8-2\sqrt{10+2\sqrt{5}}+}2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(A^2=16+2\left[64-4\left(10+2\sqrt{5}\right)\right]\)
\(A^2=16+128-8\left(10+2\sqrt{5}\right)\)
\(A^2=144-80-16\sqrt{5}\)
\(A^2=64-16\sqrt{5}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2.\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)
\(=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}=16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=2+2.\sqrt{2}.\sqrt{10}+10\)
\(=\left(\sqrt{2}+\sqrt{10}\right)^2\)
=> \(A=\sqrt{2}+\sqrt{10}\)
Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath
Bạn tham khảo link này!
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{1+2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{1-2\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=1+\sqrt{5}-\left(1-\sqrt{5}\right)=1+\sqrt{5}-1+\sqrt{5}=2\sqrt{5}\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)
c. Ta có: C+E=\(\sqrt{45+\sqrt{2009}}+\sqrt{45-\sqrt{2009}}=\sqrt{\left(\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{41}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{49}{2}}-\sqrt{\dfrac{41}{2}}\right)^2}=\dfrac{7}{\sqrt{2}}+\dfrac{\sqrt{41}}{\sqrt{2}}+\dfrac{7}{\sqrt{2}}-\dfrac{\sqrt{41}}{\sqrt{2}}=\dfrac{2.7}{\sqrt{2}}=7\sqrt{2}\)
=> đpcm.
a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)
b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
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)
haha