so sánh:\(\sqrt{\frac{3\sqrt{5}+5}{3\sqrt{5}-5}}\)với \(3+\sqrt{5}\)
Những câu hỏi liên quan
so sánh: \(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
và \(2+\sqrt{5}\)
Đặt:
\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(\left|1+\sqrt{5}\right|+\left|\sqrt{5}-1\right|\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(1+\sqrt{5}+\sqrt{5}-1\right)\)
\(A=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
Ta có: \(A^2=\left(\sqrt{10}\right)^2=10\)
\(B=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)
Mà: \(4\sqrt{5}>1\)
Nên: \(A^2< B^2\)
\(\Rightarrow A< B\)
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Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}+1+\sqrt{5}-1\right)=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
=>A^2=(căn 10)^2=10=9+1
Đặt B=2+căn 5
=>B^2=(2+căn 5)^2=9+4căn 5
1<4căn 5
=>9+1<9+4căn 5
=>A^2<B^2
=>A<B
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Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(\Rightarrow A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=6+2\sqrt{9-5}=6+2.2=10\)
\(B=2+\sqrt{5}\Rightarrow B^2=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)
\(>9+1=10=A^2\)
\(\Rightarrow B^2>A^2\Rightarrow B>A\)
Vậy, B>A
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\(A=\frac{3-\sqrt{6+\sqrt{6+\sqrt{6}}}}{3-\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}\) so sánh A với 5
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
so sánh
\(\frac{\sqrt{5}+1}{5\sqrt{10-2\sqrt{5}}}\) và \(\frac{\sqrt{3}}{6}\)
So sánh A và B biết: A=\(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\) ; B=\(\sqrt{4-\sqrt{7}}\)
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
Ta có:
\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)
\(=\dfrac{4\sqrt{2}}{\sqrt{5}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)
Làm câu S tương tự như này rồi đối chiếu kết quả nha
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So sánh :
- 10 và \(-2\sqrt{31}\)
\(2\sqrt{3}\) - 5 và \(\sqrt{5}\) - 4
2 + \(\sqrt{5}\) và 3 + \(\sqrt{2}\)
So sánh:1,2-sqrt{2}vafrac{1}{2}2, 2sqrt{3}-5vasqrt{3}-43, sqrt{3}-3sqrt{2}va-4sqrt{3}+5sqrt{2}4,1-sqrt{3}vasqrt{2}-sqrt{6}5, sqrt{4sqrt{5}}vasqrt{5sqrt{3}}6, sqrt{sqrt{6}-sqrt{5}}-sqrt{sqrt{3}-sqrt{2}}va..07, -2sqrt{frac{1}{2}sqrt{5}}va-3sqrt{frac{1}{3}sqrt{2}}
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So sánh:
1,\(2-\sqrt{2}va\frac{1}{2}\)
2, \(2\sqrt{3}-5va\sqrt{3}-4\)
3, \(\sqrt{3}-3\sqrt{2}va-4\sqrt{3}+5\sqrt{2}\)
4,\(1-\sqrt{3}va\sqrt{2}-\sqrt{6}\)
5, \(\sqrt{4\sqrt{5}}va\sqrt{5\sqrt{3}}\)
6, \(\sqrt{\sqrt{6}-\sqrt{5}}-\sqrt{\sqrt{3}-\sqrt{2}}va..0\)
7, \(-2\sqrt{\frac{1}{2}\sqrt{5}}va-3\sqrt{\frac{1}{3}\sqrt{2}}\)
1) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
2) \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(A=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=8-\frac{3+\sqrt{5}}{2}=\frac{13-\sqrt{5}}{2}\)
\(B=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{8+2\sqrt{15}+8-2\sqrt{15}}{2}=8\)
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So sánh \(\sqrt{3+\sqrt{20}}\) và \(\sqrt{5+\sqrt{5}}\)
