Ta có :

\(A=3+3^2+3^3+3^4+3^5+3^6+...+3^{58}+3^{59}+3^{60}\)

\(\Rightarrow A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(\Rightarrow A=3.\left(3^0+3^1+3^2\right)+3^4.\left(3^0+3^1+3^2\right)+...+3^{58}.\left(3^0+3^1+3^2\right)\)

\(\Rightarrow A=3.13+3^4.13+3^7.13+..+3^{58}.13\)

\(\Rightarrow A=\left(3+3^4+...+3^{58}\right).13⋮13\)

\(A=5+5^2+5^3+5^4+...+5^{11}+5^{12}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{11}+5^{12}\right)\)

\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{10}\left(5+5^2\right)\)

\(=30\left(1+5^2+...+5^{10}\right)⋮30\)

31¹¹<32¹¹=2⁵⁵

17¹⁴>16¹⁴=2⁵⁶

=>Vì 2⁵⁵<2⁵⁶ nên 31¹¹<17¹⁴.

a. f(\(\dfrac{-1}{2}\)) = \(4.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)-2\) 

               = \(4.\dfrac{1}{4}-\left(\dfrac{-3}{2}\right)-\dfrac{4}{2}\)

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

               = \(\dfrac{1}{2}\)

4*cos(pi/6-a)*sin(pi/3-a)

=4*(cospi/6*cosa+sinpi/6*sina)*(sinpi/3*cosa-sina*cospi/3)

=4*(căn 3/2*cosa+1/2*sina)*(căn 3/2*cosa-1/2*sina)

=4*(3/4*cos^2a-1/4*sin^2a)

=3cos^2a-sin^2a

=3(1-sin^2a)-sin^2a

=3-4sin^2a

=>m=3; n=-4

m^2-n^2=-7

Ta có:

\(\dfrac{1}{cos^2x-sin^2x}+\dfrac{2tanx}{1-tan^2x}=\dfrac{1}{cos2x}+tan2x=\dfrac{1}{cos2x}+\dfrac{sin2x}{cos2x}=\dfrac{1+sin2x}{cos2x}=\dfrac{cos2x}{1-sin2x}\)

\(\Rightarrow P=a+b=2+1=3\)

5: \(=\dfrac{1}{2}\cdot10-\dfrac{1}{2}=\dfrac{1}{2}\cdot9=\dfrac{9}{2}\)

Bài 9:

\(\widehat{B}=\widehat{HAC}=60^0\)