\(VT=\Pi\left(1+1+\frac{a}{b}\right)^{\alpha}\ge\Pi\left(3\sqrt[3]{\frac{a}{b}}\right)^{\alpha}=\Pi\left[3^a\sqrt[3]{\frac{a^{\alpha}}{b^{\alpha}}}\right]=3^{3a}\)?!?
Mình làm sai ak?
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rút gọn
a)A=\(\frac{1+2cos\alpha.sin\alpha}{cos^2\alpha-sin^2\alpha}\)
b)B=\(\left(1+\cot^2\alpha\right)\left(1-sin^2\alpha\right)\)-\(\left(1+\cot^2\alpha\right)\left(1-\cos^2\alpha\right)\)
c)C=\(\sin^6\alpha+\cos^6\alpha\)+\(3\sin^2\alpha.cos^2\alpha\)
14 tháng 1 2022 lúc 9:27
Đố: Cho Delta ABC, biết BCa,ACb,ABc,widehat{A}alpha,widehat{B}beta,widehat{C}gamma chứng minh:a)frac{a}{sinalpha}frac{b}{sinbeta}frac{c}{singamma} b) a^2b^2+c^2-2bccosalphac) frac{a-b}{a+b}frac{tanleft[frac{1}{2}left(alpha-betaright)right]}{tanleft[frac{1}{2}left(alpha+betaright)right]} d) Biết sfrac{a+b+c}{2}. Chứng minh frac{cotfrac{alpha}{2}}{s-a}frac{cotfrac{beta}{2}}{s-b}frac{cotfrac{gamma}{2}}{s-c}
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Đố: Cho \(\Delta ABC\), biết \(BC=a,AC=b,AB=c,\widehat{A}=\alpha,\widehat{B}=\beta,\widehat{C}=\gamma\) chứng minh:
a)\(\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}\) b) \(a^2=b^2+c^2-2bc\cos\alpha\)
c) \(\frac{a-b}{a+b}=\frac{\tan\left[\frac{1}{2}\left(\alpha-\beta\right)\right]}{\tan\left[\frac{1}{2}\left(\alpha+\beta\right)\right]}\)
d) Biết \(s=\frac{a+b+c}{2}\). Chứng minh \(\frac{\cot\frac{\alpha}{2}}{s-a}=\frac{\cot\frac{\beta}{2}}{s-b}=\frac{\cot\frac{\gamma}{2}}{s-c}\)
M^2left(sqrt{x}+sqrt{2y}right)^2left(frac{1}{_{sqrt{alpha}}}.sqrt{alpha x}+sqrt{2y}right)^2 left(frac{1}{alpha}+1right)left(alpha x+2yright)Rightarrow M^4leleft(frac{1}{alpha}+1right)^2left(alpha x+2yright)^2leleft(frac{1}{alpha}+1right)^2left(alpha^2+4right)left(x^2+y^2right)left(frac{1}{alpha}+1right)^2left(alpha^2+4right)Dấu bằng xảy ra hept{begin{cases}frac{alpha x}{frac{1}{alpha}}frac{2y}{1}frac{alpha}{x}frac{2}{y}end{cases}}Rightarrowhept{begin{cases}alpha^2x2yalphafrac{2x}{y}end{cases}R...
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\(M^2=\left(\sqrt{x}+\sqrt{2y}\right)^2=\left(\frac{1}{_{\sqrt{\alpha}}}.\sqrt{\alpha x}+\sqrt{2y}\right)^2< =\left(\frac{1}{\alpha}+1\right)\left(\alpha x+2y\right)\)
\(\Rightarrow M^4\le\left(\frac{1}{\alpha}+1\right)^2\left(\alpha x+2y\right)^2\le\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)\left(x^2+y^2\right)=\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)\)
Dấu bằng xảy ra => \(\hept{\begin{cases}\frac{\alpha x}{\frac{1}{\alpha}}=\frac{2y}{1}\\\frac{\alpha}{x}=\frac{2}{y}\end{cases}}\Rightarrow\hept{\begin{cases}\alpha^2x=2y\\\alpha=\frac{2x}{y}\end{cases}\Rightarrow\hept{\begin{cases}\frac{\alpha^2}{2}=\frac{y}{x}\\\frac{\alpha}{2}=\frac{x}{y}\end{cases}}}\Rightarrow\frac{\alpha^2}{2}=\frac{1}{\frac{\alpha}{2}}\Rightarrow\alpha=\sqrt[3]{4}\)
Suy ra max = \(\sqrt[4]{\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)}\) với \(\alpha=\sqrt[3]{4}\)
Cho \(\sin\alpha+\cos\alpha=\frac{\sqrt{6}}{2},a\in\left(0;\frac{\pi}{4}\right)\)
Tính giá trị biểu thức: \(P=\cos\left(\alpha+\frac{\pi}{4}\right)+\sqrt{2\left(1-\sin\alpha\cos\alpha+\sin\alpha-\cos\alpha\right)}\)
Chứng minh:
a)\(\cos^4\alpha-sin^4\alpha=2cos^2\alpha-1\)
b)\(\frac{cos\alpha}{1-sin\alpha}=\frac{1+sin\alpha}{cos\alpha}\)
c)\(\frac{\left(sin\alpha+cos\alpha\right)^2-\left(sin\alpha-cos\alpha\right)^2}{sin\alpha.cos\alpha}=4\)
Mình cần gấp!!!
16 tháng 10 2015 lúc 19:11
Tính :
\(B=\frac{\sin^2\alpha.\cos\left(\frac{\alpha}{2}\right)-\cot\left(\frac{\alpha}{3}\right)}{\frac{1}{\sqrt{2}}\sin\alpha+\sqrt{2}\tan\left(\frac{\alpha}{2}\right)}\) với \(\tan\alpha=\frac{\sin^267^o23'.\cos25^o41'}{\sin45^o16'+\cos^267^o29'}\text{ và }0^o
CMR: \(\frac{\sin^2\alpha}{\cos\alpha\left(1+\tan\alpha\right)}-\frac{\cos^2\alpha}{\sin\alpha\left(1+\cot\alpha\right)}=\sin\alpha-\cos\alpha\)
CM các biểu thức sau ko phụ thuộc vào giá trị góc alpha(0 alpha 90)Asin^6alpha+cos^6alpha+3sin^2alpha.cos^2alpha
Bleft(cosalpha-sinalpharight)^2+left(cosalpha+sinalpharight)^2Cfrac{left(cosalpha-sinalpharight)^2-left(cosalpha+sinalpharight)^2}{sinalpha.cosalpha}các nm làm ơn giải giúp mk nhémk camon nheii lắm ạ !!!! ^_^ ^_^
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CM các biểu thức sau ko phụ thuộc vào giá trị góc \(\alpha\)(0< \(\alpha\)< 90')
\(A=\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha
\)
\(B=\left(\cos\alpha-\sin\alpha\right)^2+\left(\cos\alpha+\sin\alpha\right)^2\)
\(C=\frac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha+\sin\alpha\right)^2}{\sin\alpha.\cos\alpha}\)
các nm làm ơn giải giúp mk nhé
mk camon nheii lắm ạ !!!! ^_^ ^_^
Cho góc nhọn alpha. Tính giá trị biểu thức:a) Aleft(sinalpha+cosalpharight)^2+left(sinalpha-cosalpharight)^2b) Bsin^4alphaleft(1+2cos^2alpharight)+cos^4alphaleft(1+2sin^2alpharight)c) Csin^6alpha+cos^6alpha+3sin^2alpha.cos^2alphad)
Dleft(3sinalpha+4cosalpharight)^2+left(4sinalpha-3cosalpharight)^2
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Cho góc nhọn \(\alpha\). Tính giá trị biểu thức:
a) \(A=\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
b) \(B=\sin^4\alpha\left(1+2\cos^2\alpha\right)+\cos^4\alpha\left(1+2\sin^2\alpha\right)\)
c) \(C=\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha\)
d)\( D=\left(3\sin\alpha+4\cos\alpha\right)^2+\left(4\sin\alpha-3\cos\alpha\right)^2\)