Mẫu vài bài thôi nhé!
\(1.C^0_n-2C^1_n+A^2_n=109\)
\(ĐK:\left\{{}\begin{matrix}n\in N^{\circledast}\\n\ge2\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1-2n+n\left(n-1\right)=109\\ \Leftrightarrow1-2n+n^2-n-109=0\\ \Leftrightarrow n^2-3n-108=0\\ \Leftrightarrow\left[{}\begin{matrix}n=12\left(N\right)\\n=-9\left(L\right)\end{matrix}\right.\)
Vậy \(n=12\)
\(2.A^2_n-C^{n-1}_n< 8\)
\(ĐK:\left\{{}\begin{matrix}n\in N^{\circledast}\\n\ge2\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow n\left(n-1\right)-\dfrac{n!}{\left(n-1\right)!}< 8\\ \Leftrightarrow n\left(n-1\right)-\dfrac{n\left(n-1\right)!}{\left(n-1\right)!}< 8\\ \Leftrightarrow n^2-n-n< 8\\ \Leftrightarrow n^2-2n-8< 0\\ \Leftrightarrow-2< n< 4\)
So với điều kiện: \(\Rightarrow\left[{}\begin{matrix}n=2\\n=3\end{matrix}\right.\)
Vậy \(n=2\) hoặc \(n=3\)
Mẫu vài bài thôi nhé!
1.C0n−2C1n+A2n=1091.Cn0−2Cn1+An2=109
ĐK:{n∈N⊛n≥2ĐK:{n∈N⊛n≥2
(1)⇔1−2n+n(n−1)=109⇔1−2n+n2−n−109=0⇔n2−3n−108=0⇔[n=12(N)n=−9(L)(1)⇔1−2n+n(n−1)=109⇔1−2n+n2−n−109=0⇔n2−3n−108=0⇔[n=12(N)n=−9(L)
Vậy n=12n=12
2.A2n−Cn−1n<82.An2−Cnn−1<8
ĐK:{n∈N⊛n≥2ĐK:{n∈N⊛n≥2
