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a) \(log_29\cdot log_34=4\)
b) \(log_{25}\cdot\dfrac{1}{\sqrt{5}}=-\dfrac{1}{4}\)
c) \(log_23\cdot log_9\sqrt{5}\cdot log_54=\dfrac{1}{2}\)
Ta có:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3 + 3\sqrt 2 }}{6}\)
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} = - \frac{{3 + \sqrt 6 }}{6}\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
\(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}=\dfrac{a^2\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{4}{5}}}{a^{\dfrac{1}{4}}}=\dfrac{a^{\dfrac{47}{15}}}{a^{\dfrac{1}{4}}}=a^{\dfrac{173}{60}}\)
\(\Rightarrow log_a\left(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}\right)=log_a\left(a^{\dfrac{173}{60}}\right)=\dfrac{173}{60}\)
\(a^{2log_a\left(\dfrac{\sqrt{105}}{30}\right)}=a^{log_a\left(\dfrac{7}{60}\right)}=\dfrac{7}{60}\)
Vậy \(B=\dfrac{173}{60}+\dfrac{7}{60}=\dfrac{180}{60}=3\)
a: \(=3\cdot3^{\dfrac{1}{2}}\cdot3^{\dfrac{1}{.4}}\cdot3^{\dfrac{1}{8}}=3^{1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}}=3^{\dfrac{15}{16}}\)
b: \(=\sqrt{a\cdot\sqrt{a\cdot a^{\dfrac{1}{2}}}}\)
\(=\sqrt{a\cdot\sqrt{a^{\dfrac{3}{2}}}}=\sqrt{a\cdot a^{\dfrac{3}{4}}}=\sqrt{a^{\dfrac{7}{4}}}=a^{\dfrac{7}{4}\cdot\dfrac{1.}{2}}=a^{\dfrac{7}{8}}\)
c: \(=\dfrac{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}}}{\left(a^{\dfrac{1}{5}}\right)^3\cdot a^{\dfrac{2}{5}}}=\dfrac{a^{\dfrac{13}{12}}}{a}=a^{\dfrac{1}{12}}\)
\(a,\sqrt{2^3}=2^{\dfrac{3}{2}}\\ b,\sqrt[5]{\dfrac{1}{27}}=\sqrt[5]{3^{-3}}=3^{-\dfrac{3}{5}}\\ c,\left(\sqrt[5]{a}\right)^4=\sqrt[5]{a^4}=a^{\dfrac{4}{5}}\)
\(a,\sqrt{42}=\sqrt{3\cdot14}>\sqrt{3\cdot12}=6\\ \sqrt[3]{51}=\sqrt[3]{17}< \sqrt[3]{3\cdot72}=6\\ \Rightarrow\sqrt{42}>\sqrt[3]{51}\\ b,16^{\sqrt{3}}=4^{2\sqrt{3}}\\ 18>12\Rightarrow3\sqrt{2}>2\sqrt{3}\Rightarrow4^{3\sqrt{2}}>4^{2\sqrt{3}}\\ \Rightarrow4^{3\sqrt{2}}>16^{\sqrt{3}}\)
\(c,\left(\sqrt{16}\right)^6=16^3=4^6=4^2\cdot4^4=4^2\cdot16^2\\ \left(\sqrt[3]{60}\right)^6=60^2=4^2\cdot15^2\\ 4^2\cdot16^2>4^2\cdot15^2\Rightarrow\sqrt{16}>\sqrt[3]{60}\Rightarrow0,2^{\sqrt{16}}< 0,2^{\sqrt[3]{60}}\)
a: \(\sqrt[4]{\left(-\dfrac{4}{5}\right)^4}=\left|-\dfrac{4}{5}\right|=\dfrac{4}{5}\)
b: \(\dfrac{\sqrt{4}}{\sqrt{5}}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)
c: \(\left(\sqrt[3]{9}\right)^2=\left(9^{\dfrac{1}{3}}\right)^2=9^{\dfrac{2}{3}}\)
d: \(\sqrt[5]{\sqrt{a}}=\sqrt[5]{a^{\dfrac{1}{2}}}=a^{\dfrac{1}{2}\cdot\dfrac{1}{5}}=a^{\dfrac{1}{10}}\)
e: \(\sqrt[3]{2^6}=\sqrt[3]{\left(2^2\right)^3}=2^2=4\)
\({\left[ {{{\left( {\frac{1}{3}} \right)}^2}} \right]^{\frac{1}{4}}}.{\left( {\sqrt 3 } \right)^5} = {\left( {\frac{1}{3}} \right)^{2.\frac{1}{4}}}.{\left( {{3^{\frac{1}{2}}}} \right)^5} = {\left( {{3^{ - 1}}} \right)^{\frac{1}{2}}}{.3^{\frac{1}{2}.5}} = {3^{ - \frac{1}{2}}}{.3^{\frac{5}{2}}} = {3^{ - \frac{1}{2} + \frac{5}{2}}} = {3^2} = 9\)
Chọn D.
a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)
\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)
b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)
a) \(\sqrt[4]{\dfrac{1}{16}}=\dfrac{1}{2}\)
b) \(\left(\sqrt[6]{8}\right)^2=\sqrt[\dfrac{6}{2}]{8}=\sqrt[3]{8}=2\)
c) \(\sqrt[4]{3}\cdot\sqrt[4]{27}=\sqrt[4]{3\cdot27}=\sqrt[4]{81}=3\)