Gọi nữ là a, nam là b

Theo đề bài ta có:

\(\left\{{}\begin{matrix}a-b=38\\b=\dfrac{4}{5}a\end{matrix}\right.\\ \Rightarrow a-\dfrac{4}{5}a=38\Rightarrow a=190\\ \Rightarrow b=\dfrac{4}{5}a=\dfrac{4}{5}.190=152\)

Vậy nữ có 190 học sinh và nam có 152 học sinh

1.C

2.B

C

 \(tana.cota=1\Rightarrow tana=\dfrac{1}{cota}=\dfrac{1}{\dfrac{40}{9}}=\dfrac{9}{40}\)

\(1+tan^2a=\dfrac{1}{cos^2a}=1+\left(\dfrac{9}{40}\right)^2=\dfrac{1681}{1600}\\ \Rightarrow cos^2a=\dfrac{1600}{1681}\\ \Rightarrow cosa=\dfrac{40}{41}\)

\(1+cot^2a=\dfrac{1}{sin^2a}=1+\left(\dfrac{40}{9}\right)^2=\dfrac{1681}{81}\\ \Rightarrow sin^2a=\dfrac{81}{1681}\\ \Rightarrow sina=\dfrac{9}{41}\)

Bài 1:

\(\left|e_c\right|=\left|\dfrac{1,2-0,6}{60}\right|=0,01\) (V)

Bài 3:

\(L=4\pi.10^{-7}.N^2.\dfrac{S}{l}\)

   \(=4\pi.10^{-7}.1000^2.\dfrac{10^{-3}}{0,5}\)

   \(=2,512.10^{-3}\left(H\right)\)

\(\dfrac{A}{B}=\dfrac{57}{86}\Rightarrow86A-57B=0\left(1\right)\)

\(\dfrac{B-10}{A+5}=\dfrac{3}{2}\Rightarrow2B-20=3A+15\Rightarrow3A-2B=-35\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}A=1995\\B=3010\end{matrix}\right.\)

\(B-A=3010-1995=1015\)

\(C_{14}^k+C_{14}^{k+2}=2C_{14}^{k+1}\)

chứng minh rằng: P_n = (n-1)P_n-1 + (n-2)P_n-2 + ... + 2P₂ + P₁ +1, với n∈N,n≥2