Bài tập 9 trang 105 SGK Toán 11 NC

a)

\(\begin{array}{l}
{u_1} = \frac{{{{2.1}^2} - 3}}{1} =  - 1\\
{u_2} = \frac{{{{2.2}^2} - 3}}{2} = \frac{5}{2}\\
{u_3} = \frac{{{{2.3}^2} - 3}}{3} = 5\\
{u_4} = \frac{{{{2.4}^2} - 3}}{4} = \frac{{29}}{4}\\
{u_5} = \frac{{{{2.5}^2} - 3}}{5} = \frac{{47}}{5}
\end{array}\)

b)

\(\begin{array}{l}
{u_1} = {\sin ^2}\frac{\pi }{4} + \cos \frac{{2\pi }}{3} = \frac{1}{2} - \frac{1}{2} = 0\\
{u_2} = {\sin ^2}\frac{\pi }{2} + \cos \frac{{4\pi }}{3} = 1 - \frac{1}{2} = \frac{1}{2}\\
{u_3} = {\sin ^2}\frac{{3\pi }}{4} + \cos 2\pi  = \frac{1}{2} + 1 = \frac{3}{2}\\
{u_4} = {\sin ^2}\pi  + \cos \frac{{8\pi }}{3} = \cos \left( {2\pi  + \frac{{2\pi }}{3}} \right) =  - \frac{1}{2}\\
{u_5} = {\sin ^2}\frac{{5\pi }}{4} + \cos \frac{{10\pi }}{3} = \frac{1}{2} - \frac{1}{2} = 0
\end{array}\)

c)

\(\begin{array}{l}
{u_1} = {\left( { - 1} \right)^1}.\sqrt {{4^1}}  =  - 2\\
{u_2} = {\left( { - 1} \right)^2}.\sqrt {{4^2}}  = 4\\
{u_3} = {\left( { - 1} \right)^3}.\sqrt {{4^3}}  =  - 8\\
{u_4} = {\left( { - 1} \right)^4}.\sqrt {{4^4}}  = 16\\
{u_5} = {\left( { - 1} \right)^5}.\sqrt {{4^5}}  =  - 32
\end{array}\)

-- Mod Toán 11 HỌC247