Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath
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15 tháng 6 2019 lúc 7:30
Chứng minh :
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
Chứng minh \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
Chứng minh rằng:
8 + 2\(\sqrt{10+2\sqrt{5}}\)+ 8 - 2\(\sqrt{10+2\sqrt{5}}\)= \(\sqrt{2}\)+ \(\left(\sqrt{5}+1\right)\)
Giúp mình với. Cảm ơn nhiều!!~
a,\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}=\dfrac{8}{1-\sqrt{5}}\)
b,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
Chứng minh các đẳng thức sau:
a)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=2\sqrt{5}\)
b)\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=4\sqrt{2}\)
c)\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}=0\)
dfrac{sqrt{3-sqrt{5}}left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}}dfrac{10+2sqrt{10}}{sqrt{5}+sqrt{2}}+dfrac{8}{1-sqrt{5}}dfrac{5+sqrt{7}}{9-sqrt{23+8sqrt{7}}}+dfrac{5-sqrt{7}}{2+sqrt{16+6sqrt{7}}}dfrac{1}{sqrt{2}+sqrt{2+sqrt{3}}}+dfrac{1}{sqrt{2}-sqrt{2+sqrt{3}}}
Đọc tiếp
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)+\(\dfrac{8}{1-\sqrt{5}}\)
\(\dfrac{5+\sqrt{7}}{9-\sqrt{23+8\sqrt{7}}}\)+\(\dfrac{5-\sqrt{7}}{2+\sqrt{16+6\sqrt{7}}}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)+\(\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
sqrt{4+sqrt{10+2sqrt{5}}}+sqrt{4-sqrt{10+2sqrt{5}}}left(sqrt{4+sqrt{10+2sqrt{5}}}+sqrt{4-sqrt{10+2sqrt{5}}}right)^24+sqrt{10+2sqrt{5}}+4-sqrt{10+2sqrt{5}}+2sqrt{left(4+sqrt{10+2sqrt{5}}right).left(4-sqrt{10+2sqrt{5}}right)}8+2sqrt{16-10-2sqrt{5}}8+2sqrt{6-2sqrt{5}}8+2sqrt{left(sqrt{5}-1right)^2}8+2left(sqrt{5}-1right)8+2sqrt{5}-26+2sqrt{5}left(sqrt{5+1}right)^2sqrt{left(sqrt{5}+1right)^2}|sqrt{5}+1|sqrt{5}+1text{thư ngu như chó}
Đọc tiếp
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right).\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(8+2\sqrt{16-10-2\sqrt{5}}\)
\(8+2\sqrt{6-2\sqrt{5}}\)
\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(8+2\left(\sqrt{5}-1\right)\)
\(8+2\sqrt{5}-2\)
\(6+2\sqrt{5}\)
\(\left(\sqrt{5+1}\right)^2\)
\(\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(|\sqrt{5}+1|\)
\(\sqrt{5}+1\)
\(\text{thư ngu như chó}\)
chứng minh rằng : \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5}\)