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18 tháng 8 2023

a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)

b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)

c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)

\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)

a: \(A=\dfrac{x^5}{x^3}\cdot\dfrac{y^{-2}}{y}=x^2\cdot y^{-1}=\dfrac{x^2}{y}\)

b: \(B=\dfrac{x^2\cdot y^{-3}}{x^3\cdot y^{-12}}=\dfrac{x^2}{x^3}\cdot\dfrac{y^{-3}}{y^{-12}}=\dfrac{1}{x}\cdot y^{-3+12}=\dfrac{y^9}{x}\)

 

23 tháng 8 2023

a) \(A=\dfrac{x^5y^{-2}}{x^3y}=\dfrac{x^5}{x^3}.\dfrac{1}{y^{2-1}}=x^{5-3}y^{-1}=x^2y^{-1}\).

b) \(B=\dfrac{x^2y^{-3}}{\left(x^{-1}y^4\right)^{-3}}=\dfrac{x^2y^{-3}}{x^3y^{-12}}=x^{2-3}y^{-3-\left(-12\right)}=\dfrac{1}{xy^9}\)

HQ
Hà Quang Minh
Giáo viên
24 tháng 8 2023

\(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}=\dfrac{a^2\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{4}{5}}}{a^{\dfrac{1}{4}}}=\dfrac{a^{\dfrac{47}{15}}}{a^{\dfrac{1}{4}}}=a^{\dfrac{173}{60}}\)

\(\Rightarrow log_a\left(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}\right)=log_a\left(a^{\dfrac{173}{60}}\right)=\dfrac{173}{60}\)

\(a^{2log_a\left(\dfrac{\sqrt{105}}{30}\right)}=a^{log_a\left(\dfrac{7}{60}\right)}=\dfrac{7}{60}\)

Vậy \(B=\dfrac{173}{60}+\dfrac{7}{60}=\dfrac{180}{60}=3\)

HQ
Hà Quang Minh
Giáo viên
25 tháng 8 2023

\(a,a^{\dfrac{3}{5}}\cdot a^{\dfrac{1}{2}}:a^{-\dfrac{2}{5}}=a^{\dfrac{3}{5}+\dfrac{1}{2}-\left(-\dfrac{2}{5}\right)}=a^{\dfrac{3}{2}}\\ b,\sqrt{a^{\dfrac{1}{2}}\sqrt{a^{\dfrac{1}{2}}\sqrt{a}}}\\ =\sqrt{a^{\dfrac{1}{2}}\sqrt{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}}}}\\ =\sqrt{a^{\dfrac{1}{2}}\sqrt{a}}\\ =\sqrt{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}}}\\ =\sqrt{a}\)

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: \(A=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{2}}+y^{\dfrac{1}{3}}\cdot x^{\dfrac{1}{2}}}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}\left(x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}\right)}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}=\left(xy\right)^{\dfrac{1}{3}}\)

b: \(B=\dfrac{x^{3+\sqrt{3}}}{y^2}\cdot\dfrac{x^{-\sqrt{3}-1}}{y^{-2}}=\dfrac{x^{3+\sqrt{3}-\sqrt{3}-1}}{y^{2-2}}=x^2\)

\(=\dfrac{a^{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}{a^{\left(\sqrt{5}-1\right)+\left(3-\sqrt{5}\right)}}=\dfrac{a}{a^{\sqrt{5}-1+3-\sqrt{5}}}=\dfrac{a}{a^2}=\dfrac{1}{a}\)

 

HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023

Ta có :

\(\sin \left( {a + \frac{\pi }{4}} \right) + \sin \left( {a - \frac{\pi }{4}} \right) = 2.\sin a.\cos \frac{\pi }{4} =  - \frac{2}{3}\)

Chọn C

HQ
Hà Quang Minh
Giáo viên
22 tháng 8 2023

a,Ta có: \(a^6=\left(a^{\dfrac{3}{5}}\right)^{10}=b^{10}\\ a^3b=\left(a^{\dfrac{3}{5}}\right)^5\cdot b=b^5\cdot b=b^6\\ \dfrac{a^9}{b^9}=\dfrac{\left(a^{\dfrac{3}{5}}\right)^{15}}{b^9}=\dfrac{b^{15}}{b^9}=b^6\)

b, \(log_ab=log_aa^{\dfrac{3}{5}}=\dfrac{3}{5}\\ log_a\left(a^2b^5\right)=log_a\left(a^2\cdot a^3\right)=log_a\left(a^5\right)=5\\ log_{\sqrt[5]{a}}\left(\dfrac{a}{b}\right)=5log_a\left(\dfrac{a}{a^{\dfrac{3}{5}}}\right)=5log_a\left(a^{\dfrac{2}{5}}\right)=2\)