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a.\(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right).\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)
\(=\left(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\right).\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)
\(=\left(\sqrt{3}+1-\sqrt{3}+1\right)\left(\sqrt{3}-1+\sqrt{3}+1\right)\)
\(=2.2\sqrt{3}=4\sqrt{3}\)
b.\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=\left[\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\right]^2\)
\(=\left(\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\right)^2\)
\(=\left(\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}\right)^2=\left(\sqrt{2}\right)^2=2\)
c.\(\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}=\sqrt{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{1}=1\)
b,c
\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)
=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
ta thấy A> 2
Xét A2 = 5 + \(\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5...........}}}}\)
( A2 - 5 )2 = 13 + A
<=> A4 - 10A2 - A + 12 = 0
<=> (A4 - 9A2 ) - ( A2 - 9 ) - (A - 3) = 0
<=> (A - 3) [(A + 3)(A+1)(A-1)-1] =0
Vì A> 2 => (A + 3)(A+1)( A-1)-1 > 0
Do đó A - 3 = 0 <=> A = 3
\(\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}+\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}}=\sqrt{\frac{7-3\sqrt{5}}{2}+\frac{3+\sqrt{5}}{2}}\)
\(=\sqrt{5-\sqrt{5}}\)
\(E=\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}.\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)\)
\(E=\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{5\sqrt{6}-12}{18\sqrt{2}}\)
\(E=\frac{36\sqrt{2}}{18\sqrt{6}}+\frac{12\sqrt{3}}{18\sqrt{6}}+\frac{\left(5\sqrt{6}-12\right).\sqrt{3}}{18\sqrt{3}}\)
\(E=\frac{36\sqrt{2}+12\sqrt{3}+\left(5\sqrt{6}-12\right).\sqrt{3}}{18\sqrt{6}}\)
\(E=\frac{51\sqrt{2}}{18\sqrt{6}}\)
\(E=\frac{17\sqrt{2}}{6\sqrt{6}}\)
\(E=\frac{17\sqrt{2}}{2.3\sqrt{2}.\sqrt{3}}\)
\(E=\frac{17}{\sqrt{2}.3\sqrt{2}.\sqrt{3}}\)
\(E=\frac{17}{6\sqrt{3}}\)
\(E=\frac{17\sqrt{3}}{18}\)
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
\(A=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+...}}}}=>A^2=3+\sqrt{3+\sqrt{3+\sqrt{3+...}}}\)
\(=>A^2-A=3< =>A^2-A-3=0=>\orbr{\begin{cases}A=\frac{1+\sqrt{13}}{2}\\A=\frac{1-\sqrt{13}}{2}\left(A>0=>l\right).\end{cases}}\)
Vậy \(A=\frac{1+\sqrt{13}}{2}.\)
Bạn đạt \(\sqrt{3+\sqrt{3+\sqrt{3+...}}=A}\)
Ta có ,,do đó là dãy số vô hạn nên \(A=\sqrt{3+A}\)
Từ đó tính đc giá trị xấp xỉ của A chú ý A>0