CMR: \(\sin18^0=\frac{\sqrt{5}-1}{4}\)
NL
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Chứng minh \(sin18=\frac{\sqrt{5}-1}{4}\)
Cho sin18 = \(\frac{\sqrt{5}-1}{4}\). Tính cos18, sin72, sin162, cos162, sin108, cos108, tan72
1a) 6sqrt{3} - 5sqrt{12} + 3sqrt{75} b) 2sqrt{5} - dfrac{1}{4} sqrt{80} + 7sqrt{500} c) dfrac{sin43^o}{cos47^o} + tan45o d) dfrac{tan32^o}{tan68^o} - cos30o - dfrac{sin18^o}{sin82^o}
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1a) 6\(\sqrt{3}\) - 5\(\sqrt{12}\) + 3\(\sqrt{75}\)
b) 2\(\sqrt{5}\) - \(\dfrac{1}{4}\) \(\sqrt{80}\) + 7\(\sqrt{500}\)
c) \(\dfrac{sin43^o}{cos47^o}\) + tan45o
d) \(\dfrac{tan32^o}{tan68^o}\) - cos30o - \(\dfrac{sin18^o}{sin82^o}\)
\(a,=6\sqrt{3}-10\sqrt{3}+15\sqrt{3}=11\sqrt{3}\\ b,=2\sqrt{5}-\sqrt{5}+70\sqrt{5}=71\sqrt{5}\\ c,=\dfrac{\sin43^0}{\sin43^0}+1=1+1=2\\ d,Sửa:\dfrac{\tan32^0}{\cot68^0}-\cos30^0-\dfrac{\sin18^0}{\sin82^0}=\dfrac{\tan32^0}{\tan32^0}-\dfrac{\sqrt{3}}{2}-\dfrac{\sin18^0}{\cos18^0}=1-1-\dfrac{\sqrt{3}}{2}=-\dfrac{\sqrt{3}}{2}\)
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\(e.B=\frac{3+\sqrt{x}}{4+\sqrt{x}}\left(0\le x< 1\right)\)
CMR: B < \(\frac{4}{5}\)
\(c.C=\frac{\sqrt{x}+1}{\sqrt{x}+3};D=\frac{\sqrt{x}+2}{\sqrt{x}+4}\left(x\ge0\right)\)
CMR : C<D
\(d.\frac{\sqrt{x}}{\sqrt{x}+\sqrt{x}+4}\left(x>0\right)\)
CMR : \(D< \frac{1}{4}\)
1) Cho sin(400 + x) = \(\frac{4}{5}\) và 600 < x < 900. Tính cos(700 + x)
2) Cho \(cos\left(a-\frac{b}{2}\right)=-\frac{1}{9};sinb\left(b-\frac{a}{2}\right)=-\frac{2}{3}\) (0 < b < 900 < a < 1800). Tính cos(a+b)
3) Tính A=\(\frac{1}{sin18^0}-\frac{1}{sin54^0}\)
CMR : \(\frac{\sqrt[4]{5}-1}{\sqrt{5}+1}=\sqrt[4]{\frac{3-2\sqrt[4]{5}}{3+2\sqrt[4]{5}}}\)
CMR \(\frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1}=\sqrt[4]{\frac{3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}\)
Đặt \(a=\sqrt[4]{5}\Leftrightarrow5=a^4\)
Ta cần chứng minh: \(\left(\frac{a+1}{a-1}\right)^4=\frac{3+2a}{3-2a}\)
Khai triển: \(VT=\left(\frac{a+1}{a-1}\right)^4=\frac{\left(a+1\right)^4}{\left(a-1\right)^4}\)
\(=\frac{2\left(3+2a\right).\left(1+a^2\right)}{2\left(3-2a\right).\left(1+a^2\right)}\)
\(\frac{3+2a}{3-2a}=VP\)(đpcm)
Bai 1: cho ninℕ^∗. CMR : frac{1}{2}.frac{3}{4}.frac{5}{6}.....frac{2n-1}{2n} frac{1}{sqrt{3n+1}}. nghia la be hon hoac bang nha cac banBai 2 : Cho a0;b0. CMR : frac{2sqrt{ab}}{sqrt{a}+sqrt{b}} sqrt{sqrt{ab}}Bai 3: Cho x, y, z 0 và x + y + z 1. Chứng minh rằng:sqrt{x+yz}+sqrt{y+zx}+sqrt{z+xy}1+sqrt{xy}+sqrt{yz}+sqrt{zx}
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Bai 1: cho \(n\inℕ^∗\). CMR : \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{2n-1}{2n}< =\frac{1}{\sqrt{3n+1}}\). <= nghia la be hon hoac bang nha cac ban
Bai 2 : Cho a>0;b>0. CMR : \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}< =\sqrt{\sqrt{ab}}\)
Bai 3: Cho x, y, z > 0 và x + y + z = 1. Chứng minh rằng:\(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}>=1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(3,\)Áp dụng bđt Mincopski \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)hai lần có
\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)
\(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)
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\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)
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CMR \(\sin22,\:5^0=\frac{\sqrt{3+\sqrt{2}}}{\sqrt{4+2\sqrt{2}}}\)
Kết quả này sai rồi bạn, bạn có thể kiểm tra lại bằng máy tính.
Dựng tam giác vuông cân ABC có \(AB=AC=1\); \(BC=\sqrt{2}\)
Dựng phân giác BD của góc B \(\Rightarrow\widehat{ABD}=\frac{45}{2}=22,5^0\)
Theo t/c phân giác: \(\frac{AD}{AB}=\frac{CD}{BC}\Rightarrow CD=\sqrt{2}AD\)
Mà \(AD+CD=AB\Rightarrow AD+\sqrt{2}AD=1\Rightarrow AD=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
\(BD=\sqrt{AB^2+BD^2}=\sqrt{1+\left(\sqrt{2}-1\right)^2}=\sqrt{4-2\sqrt{2}}\)
\(\Rightarrow sin22,5^0=sin\widehat{ABD}=\frac{AD}{BD}=\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}\)
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