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a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{1}=1}\)
b) \(B=\sqrt{\sqrt{3}-\sqrt{1+\sqrt{21-6\sqrt{12}}}=\sqrt{\sqrt{3}-\sqrt{1+\sqrt{\left(3-2\sqrt{3}\right)^2}}}}=\sqrt{\sqrt{3}-\sqrt{2\sqrt{3}-2}}\)c)
\(C=\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}=\frac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{2+2\sqrt{5}}{\sqrt{2}}=\sqrt{2}+\sqrt{10}=\sqrt{2}\left(\sqrt{5}+1\right)\)
a) \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(\Leftrightarrow-\left(\sqrt{3}+11\sqrt{5}+\sqrt{29}\right)\)
\(\Leftrightarrow\sqrt{637+22\sqrt{145}+2\sqrt{6\left(317+11\sqrt{145}\right)}}\)
\(\Leftrightarrow\sqrt{3}-11\sqrt{5}-\sqrt{29}\)
b) Câu hỏi của Nguyễn Trung Anh - Toán lớp 9 - Học toán với OnlineMath giống câu này!
a/ \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\sqrt{5}+1=1\)
b/ Câu hỏi của Nguyễn Trung Anh - Toán lớp 9 - Học toán với OnlineMath giống câu này.
Trả lời
\(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
Đặt \(M=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)
\(M^2=\left(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\right)^2\)
\(M^2=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)^2}{\left(\sqrt{\sqrt{5}+1}\right)^2}\)
\(M^2=\frac{\sqrt{5}+2+2\sqrt{\left(\sqrt{5}+2\right).\left(\sqrt{5}-2\right)}+\sqrt{5}-2}{\sqrt{5}+1}\)
\(M^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}\)
\(M^2=\frac{2\sqrt{5}+2}{\sqrt{5}+1}\)
\(M^2=\frac{2.\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\)
\(M^2=2\)
\(M=\sqrt{2}\)
THay M vào B ta có \(B=M-\sqrt{3-2\sqrt{2}}\)
\(B=\sqrt{2}-\sqrt{3-2\sqrt{2}}\)
\(B=\sqrt{2}-\sqrt{2-2\sqrt{2}+1}\)
\(B=\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(B=\sqrt{2}-\sqrt{2}+1\)
\(B=1\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=\sqrt{1}=1\)
\(A=\sqrt[3]{8-\sqrt{60}}+\sqrt[3]{8+\sqrt{60}}\) xem lại đề con này
\(A=\frac{2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{4+2\sqrt{3}}}{2\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
ta có: A3=\(6\sqrt{3}+10-6\sqrt{3}+10-3\sqrt[3]{\left(6\sqrt{3}+10\right)\left(6\sqrt{3}-10\right)}.\left(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\right)\)
=\(20-3.\sqrt[3]{8}.A\)=\(20-6A\)
do đó A3=20-6A↔A3+6A-20=0↔(A2+2A+10)(A-2)=0
dễ thấy A2+2A+10>0→A=2
b) giống a)
c)giống b)
\(a,\sqrt{29+12\sqrt{5}}+2\sqrt{21-8\sqrt{5}}\)
\(\sqrt{29+6\sqrt{20}}+\sqrt{84-32\sqrt{5}}\)
\(\sqrt{\sqrt{20}^2+6\sqrt{20}+3^2}+\sqrt{84-16\sqrt{20}}\)
\(\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{8^2-16\sqrt{20}+\sqrt{20}^2}\)
\(\left|\sqrt{20}+3\right|+\sqrt{\left(8-\sqrt{20}\right)^2}\)
\(\sqrt{20}+3+\left|8-\sqrt{20}\right|\)
\(\sqrt{20}+3+8-\sqrt{20}\)
\(=11\)