\(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)+\(\dfrac{8}{1-\sqrt{5}}\)
\(\dfrac{5+\sqrt{7}}{9-\sqrt{23+8\sqrt{7}}}\)+\(\dfrac{5-\sqrt{7}}{2+\sqrt{16+6\sqrt{7}}}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)+\(\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
đề là rút gọn các biểu thức sau
nhờ mọi người giải giúp mình. cảm ơn mn nhìu
a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}\)
\(=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)
=2căn 5-2-2căn 5
=-2
d: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}=\sqrt{2}\)
Chứng minh :
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}+8-2\sqrt{10+2\sqrt{5}}+}2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(A^2=16+2\left[64-4\left(10+2\sqrt{5}\right)\right]\)
\(A^2=16+128-8\left(10+2\sqrt{5}\right)\)
\(A^2=144-80-16\sqrt{5}\)
\(A^2=64-16\sqrt{5}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2.\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)
\(=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}=16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=2+2.\sqrt{2}.\sqrt{10}+10\)
\(=\left(\sqrt{2}+\sqrt{10}\right)^2\)
=> \(A=\sqrt{2}+\sqrt{10}\)
Tính:
\(\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}-\sqrt{2}-\sqrt{10}\)
A = \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}-\sqrt{2}-\sqrt{10}\)
Ta có : B = \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow B^2=16-2\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)
\(=16-2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)
\(=16-2\sqrt{24-8\sqrt{5}}\)
\(=16-2\sqrt{\left(2\sqrt{5}-2\right)^2}=16-2\left(2\sqrt{5}-2\right)\)
\(=20-4\sqrt{5}\)
Vì \(8+2\sqrt{10+\sqrt{5}}>8-2\sqrt{10+2\sqrt{5}}\)
\(\Rightarrow B>0\)
\(\Rightarrow B=\sqrt{20-4\sqrt{5}}=2\sqrt{5-\sqrt{5}}\)
\(\Rightarrow A=B-\sqrt{2}-\sqrt{10}=2\sqrt{5-\sqrt{5}}-\sqrt{2}-\sqrt{10}=2\)
Chứng minh rằng
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath
Bạn tham khảo link này!
Chứng minh \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
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)
bn thang tran lm sai bước đưa ra hdt :v đúng là phải 16 - ( 10 + 2can5 )
= 16 - 10 - 2can5
a,\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}=\dfrac{8}{1-\sqrt{5}}\)
b,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right).\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(8+2\sqrt{16-10-2\sqrt{5}}\)
\(8+2\sqrt{6-2\sqrt{5}}\)
\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(8+2\left(\sqrt{5}-1\right)\)
\(8+2\sqrt{5}-2\)
\(6+2\sqrt{5}\)
\(\left(\sqrt{5+1}\right)^2\)
\(\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(|\sqrt{5}+1|\)
\(\sqrt{5}+1\)
\(\text{thư ngu như chó}\)
\(\sqrt{8+2\sqrt{10}+2\sqrt{5}}+\sqrt{8-2\sqrt{10}+2\sqrt{5}}\)
Đặt S = \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{8-2\sqrt{10-2\sqrt{5}}}\)
S2 = 8 + 2\(\sqrt{10+2\sqrt{5}}\) + \(8-2\sqrt{10+2\sqrt{5}}\) + 2\(\times\)\(\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)
= 16 + \(2\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)
= 16 + 2 \(\sqrt{64-40+8\sqrt{5}}\)
= 16 + 2\(\sqrt{20+2\times2\sqrt{5}\times2+4}\)
= 16 + 2\(\sqrt{\left(\sqrt{20}+2\right)^2}\)
= 16 + 2\(\sqrt{20}-4\)
= 12 + 2\(\sqrt{20}\)
Do S > 0 nên
S = \(\sqrt{12+2\sqrt{20}}\)= \(\sqrt{12+2\times2\sqrt{5}}\)=\(\sqrt{4\left(3+\sqrt{5}\right)}\)=\(2\sqrt{3+\sqrt{5}}\)
Vậy S = 2\(\sqrt{3+\sqrt{5}}\)
\(\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)