Câu 18.

Nhiệt lượng miếng đồng thu vào:

\(Q=mc\left(t_2-t_1\right)\)

\(\Rightarrow30400=m\cdot380\cdot\left(50-20\right)\)

\(\Rightarrow m=2,67kg\)

20: \(\lim\limits_{x\rightarrow+\infty}x^3+2x-1=\lim\limits_{x\rightarrow+\infty}\left[x^3\left(1+\dfrac{2}{x^2}-\dfrac{1}{x^3}\right)\right]\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow+\infty}x^3=+\infty\\\lim\limits_{x\rightarrow+\infty}1+\dfrac{2}{x^2}-\dfrac{1}{x^3}=1\end{matrix}\right.\)

  9C

16:C

15 D (relieved : an tâm)

17 B (however + tính từ)

$4.17+4.84-4$

$=4.17+4.84-4.1$

$=4.17+4.84-1.4$

$=4.(17+84-1)$

$=4.100$

$=400$

_________________

$17.18-18+84.18$

$=17.18-18.1+84.18$

$=17.18+18.(-1)+84.18$

$=18.(17-1+84)$

$=18.100$

$=1800$

_____________________

$75.12+75.9$

$75.12+75.9-21.25$

$=75.(12+9)-21.25$

$=75.21-21.25$

$=75.21+21.(-25)$

$=21.(75-25)$

$=21.50$

$=1050$

câu 8.D

câu 13.C

câu 15.C

câu 16 ko biết

câu 17.A

câu 20.D

Câu 10: 

Gọi \(H\) là giao điểm của \(MO\) và \(AB\).

Xét tam giác \(MAO\) vuông tại \(A\) đường cao \(AH\): 

\(\dfrac{1}{AH^2}=\dfrac{1}{MA^2}+\dfrac{1}{AO^2}\Leftrightarrow\dfrac{1}{\left(\dfrac{R\sqrt{2}}{2}\right)^2}=\dfrac{1}{MA^2}+\dfrac{1}{R^2}\Leftrightarrow MA=R\).

\(S_{MAOB}=S_{MAO}+S_{MBO}\)

\(=\dfrac{1}{2}.AO.MA+\dfrac{1}{2}.OB.MB\)

\(=\dfrac{1}{2}.R.R+\dfrac{1}{2}.R.R=R^2\)

Chọn C.