\(a,\dfrac{3^{43}+3^4}{3^{39}+1}\)

\(=\dfrac{3^4\cdot\left(3^{39}+1\right)}{3^{39}+1}\)

\(=3^4\)

\(=81\)

\(b,\dfrac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(=\dfrac{3^{10}\left(11+5\right)}{3^9.2^4}\)

\(=\dfrac{3^{10}.16}{3^9.2^4}\)

\(=\dfrac{3^{10}.2^4}{3^9.2^4}=3\)

d: \(\left(x-2\right)\left(x^2+2x+4\right)\left(x+2\right)\left(x^2-2x+4\right)\)

\(=\left(x^3-8\right)\left(x^3+8\right)\)

\(=x^6-64\)

C = \(\frac{2^{10}.13+2^{10}.65}{2^6.104}\)

    = \(\frac{2^{10}.\left(13+65\right)}{2^6.104}\)

    = \(\frac{2^{10}.78}{2^6.104}\)

    = \(\frac{2^6.2^4.78}{2^6.104}\)

    = \(\frac{2^6.16.78}{2^6.4.26}\)

    = \(\frac{2^6.4.4.26.3}{2^6.4.26}\)

    = \(4.3\)

    = \(12\)

\(a,=-x^3+3x^2+2x-6\)

\(\left(x^2-2\right)\left(-x+3\right)\)

\(=-x^3+3x^2+2x-6\)

\(A=\dfrac{1}{2}sin4x-\dfrac{1}{2}sin2x+\dfrac{1}{2}sin12x+\dfrac{1}{2}sin2x\)

\(=\dfrac{1}{2}sin4x+\dfrac{1}{2}sin12x=sin8x.cos4x\)

\(B=\dfrac{1}{2}cos5x+\dfrac{1}{2}cos3x+\dfrac{1}{2}cosx-\dfrac{1}{2}cos5x\)

\(=\dfrac{1}{2}cos3x+\dfrac{1}{2}cosx=cos2x.cosx\)

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+............+2015.2016.(2017-2014)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.............+2015.2016.2017-2014.2015.2016

3A=2015.2016.2017

A=2015.2016.2017:3

A=2015.672.2017

\(\dfrac{25}{30}=\dfrac{5}{7};\dfrac{9}{15}=\dfrac{3}{5};\dfrac{6}{10}=\dfrac{3}{5};\dfrac{10}{12}=\dfrac{5}{6}\)

tự xem cái nào giống

a) \(\dfrac{3}{5};\dfrac{5}{6}\) : phân số tối giản
\(\dfrac{25}{30}=\dfrac{25:5}{30:5}=\dfrac{5}{6}\)
\(\dfrac{9}{15}=\dfrac{9:3}{15:3}=\dfrac{3}{5}\)
\(\dfrac{10}{12}=\dfrac{10:2}{12:2}=\dfrac{5}{6}\)
\(\dfrac{6}{10}=\dfrac{6:2}{10:2}=\dfrac{3}{5}\)
b) \(\dfrac{3}{5}=\dfrac{9}{15}=\dfrac{6}{10}\) ; \(\dfrac{5}{6}=\dfrac{25}{30}=\dfrac{10}{12}\)

\(A=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{3}\)

\(A=3+3^2+3^3+...+3^{2004}\)

\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2004}\right)\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2005}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2005}\right)-\left(3+3^2+3^3+3^4+...+3^{2004}\right)\)

\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+\left(3^4-3^4\right)+...+\left(3^{2004}-3^{2004}\right)+\left(3^{2005}-3\right)\)

\(\Rightarrow2A=3^{2005}-3\)

\(\Rightarrow A=\dfrac{3^{2005}+3}{2}\)