Bài 3: Cấp số cộng

Theo giả thiết ta có 3 góc: \(\alpha;\beta=\alpha+\dfrac{\pi}{3};\gamma=\alpha+\dfrac{2\pi}{3}\).
Ta có:
\(tan\alpha.tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{\pi}{3}\right).tan\left(\alpha+\dfrac{2\pi}{3}\right)+\)\(tan\left(\alpha+\dfrac{2\pi}{3}\right).tan\alpha\)
\(=tan\alpha\left[tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{2\pi}{3}\right)\right]\)\(+tan\left(a+\dfrac{\pi}{3}\right)tan\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=tan\alpha\dfrac{sin\left(2\alpha+\pi\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{sin\left(\alpha+\dfrac{\pi}{3}\right)sin\left(\alpha+\dfrac{2\pi}{3}\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=tan\alpha\dfrac{-sin2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{cos\dfrac{\pi}{3}-cos\left(2\alpha+\pi\right)}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{-2sin^2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{\dfrac{1}{2}+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4sin^2\alpha+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4\left(1-cos^2\alpha\right)+2cos^2\alpha-1}{cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)}\)
\(=\dfrac{6cos^2\alpha-\dfrac{9}{2}}{\dfrac{1}{2}-cos2\alpha}\)
\(=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{1}{2}-\left(2cos^2\alpha-1\right)}=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{3}{2}-2cos^2\alpha}=-3\).

Bài làm

a)dãy số U: \(2,7,12,...x\)

U là cấp số cộng\(\Rightarrow\left\{{}\begin{matrix}d=u_2-u_1=7-2=5\\u_1=2\end{matrix}\right.\)

\(U_n=U_1+\left(n-1\right)d\)

=> \(n=\dfrac{U_n-U_1}{d}+1=\dfrac{x-2}{5}+1=\dfrac{\left(x+3\right)}{5}\)

\(S_n=\dfrac{n\left(U_1+U_n\right)}{2}=\dfrac{\dfrac{\left(x+3\right)}{5}\left(2+x\right)}{2}=\dfrac{\left(x+3\right)\left(x+2\right)}{2.5}=245\)

\(x^2+5x+6=2450\)

\(x^2+5x-2444=0\)

\(\Delta=5^2-4.\left(-2444\right)=9801=\)99^2

\(\left\{{}\begin{matrix}x_1=\dfrac{-5-99}{2}< 0\left(loai\right)\\x_2=\dfrac{-5+99}{2}=47\end{matrix}\right.\)

Đáp số: x=47