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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}=\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}=\dfrac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\sqrt{\left(\sqrt{3}+1\right)^2}=\dfrac{\sqrt{3}+1}{\sqrt{2}}=\dfrac{\sqrt{2}+\sqrt{6}}{2}\)

\(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}\\ =\sqrt{3+\sqrt{5-1+2\sqrt{3}}}\\ =\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{3+\sqrt{3}-1}\\ =\sqrt{2+\sqrt{3}}\)

Ta có: \(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{3+\sqrt{3}-1}\)

\(=\sqrt{2-\sqrt{3}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)

Ta có: \(A=\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)

\(=2\left(\sin^2\alpha+\cos^2\alpha\right)\)

=2

a.\(\frac{5100-10}{8160-16}\)= \(\frac{5090}{8144}\) = \(\frac{5}{8}\)
b.\(\frac{952+8}{2142+18}\) = \(\frac{960}{2160}\) = \(\frac{4}{9}\)
Chúc bạn học tốt!

\(A=\dfrac{x}{x-2}-\dfrac{x^2+x-2}{x^2-4}=\dfrac{x^2+2x-x^2-x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

\(A=\dfrac{x}{x-2}+\dfrac{x^2+x-2}{4-x^2}\left(x\ne\pm2\right).\)

\(A=\dfrac{x}{x-2}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}-\dfrac{x-1}{x-2}=\dfrac{x-x+1}{x-2}=\dfrac{1}{x-2.}\)

A= \(\dfrac{x}{x-2}+\dfrac{x^2+x-2}{4-x^2}=\dfrac{-x\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\dfrac{x^2+x-2}{\left(2-x\right)\left(2+x\right)}=\dfrac{-2x-x^2+x^2+x-2}{\left(2-x\right)\left(2+x\right)}=\dfrac{-x-2}{\left(2-x\right)\left(2-x\right)}=\dfrac{-1\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}=\dfrac{-1}{2-x}\)