Câu 11:
\(S_n=253\)
=>\(n\cdot\dfrac{\left[2u_1+\left(n-1\right)\cdot d\right]}{2}=253\)
=>\(n\cdot\dfrac{\left[2\cdot3+\left(n-1\right)\cdot4\right]}{2}=253\)
=>\(n\left(3+2n-2\right)=253\)
=>\(2n^2+n-253=0\)
=>n=11
=>Chọn D
Câu 12: \(\left\{{}\begin{matrix}u_1+u_2+u_3=13\\u_4-u_1=26\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}u_1+u_1\cdot q+u_1\cdot q^2=13\\u_1\cdot q^3-u_1=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1\left(q^2+q+1\right)=13\\u_1\left(q^3-1\right)=26\end{matrix}\right.\)
=>\(\dfrac{q^3-1}{q^2+q+1}=\dfrac{26}{13}=2\)
=>q-1=2
=>q=3
\(u_1=\dfrac{13}{3^2+3+1}=\dfrac{13}{9+4}=1\)
\(S_8=\dfrac{u_1\left(1-q^8\right)}{1-q}=\dfrac{1\cdot\left(1-3^8\right)}{1-3}=\dfrac{3^8-1}{3-1}=3280\)
=>Chọn D
Câu 14:
a: Sai vì \(sin\left(\dfrac{\Omega}{6}\right)=\dfrac{1}{2}\)
b: \(sin2x=-\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x=-\dfrac{\Omega}{6}+k2\Omega\\2x=\Omega+\dfrac{\Omega}{6}+k2\Omega=\dfrac{7}{6}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{\Omega}{12}+k\Omega\\x=\dfrac{7}{12}\Omega+k\Omega\end{matrix}\right.\)
\(x\in\left(0;\Omega\right)\)
=>\(\left[{}\begin{matrix}-\dfrac{\Omega}{12}+k\Omega\in\left(0;\Omega\right)\\\dfrac{7}{12}\Omega+k\Omega\in\left(0;\Omega\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k-\dfrac{1}{12}\in\left(0;1\right)\\k+\dfrac{7}{12}\in\left(0;1\right)\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}k\in\left(\dfrac{1}{12};\dfrac{13}{12}\right)\\k\in\left(-\dfrac{7}{12};\dfrac{5}{12}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k=1\\k=0\end{matrix}\right.\)
=>Trên đoạn (0;pi) phương trình có 2 nghiệm
=>Sai
c: Khi k=1 thì \(x=-\dfrac{\Omega}{12}+1\cdot\Omega=\dfrac{5}{12}\Omega\)
Khi k=0 thì \(x=\dfrac{7}{12}\Omega+0\Omega=\dfrac{7}{12}\Omega\)
Tổng các nghiệm trong khoảng (0;pi) là: \(\dfrac{5}{12}\Omega+\dfrac{7}{12}\Omega=\dfrac{12}{12}\Omega=\Omega\)
=>Sai
d: Sai
