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a: \(=\left(4+\dfrac{3}{4}+\dfrac{1}{8}+3+\dfrac{1}{12}\right)+\left(-0.37-1.28-2.5\right)=\dfrac{191}{24}-\dfrac{83}{20}=\dfrac{457}{120}\)
b: \(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)
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a) \(A=sin^254^o.tan17^o.tan73^o+cos54^o.sin34^o\)
\(=sin^254^o.tan17^o.cot17^o+cos54^o.cos54^o\)
\(=sin^254^o+cos^254^o=1\)
b) \(B=\dfrac{cosa+sina}{sina-cosa}=\dfrac{\dfrac{cosa}{sina}+1}{1-\dfrac{cosa}{sina}}=\dfrac{cota+1}{1-cota}\)
\(=\dfrac{1+\sqrt{3}}{1-\sqrt{3}}=\dfrac{\left(1+\sqrt{3}\right)^2}{1-3}=\dfrac{1+2\sqrt{3}+3}{-2}=-2-\sqrt{3}\)
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d: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: \(\Leftrightarrow1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{5x+1}-\dfrac{1}{5x+6}=\dfrac{2010}{2011}\)
\(\Leftrightarrow1-\dfrac{1}{5x+6}=\dfrac{2010}{2011}\)
=>1/(5x+6)=1/2011
=>5x+6=2011
=>5x=2005
hay x=401