Câu 18.

Nhiệt lượng miếng đồng thu vào:

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

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

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

Ta có: \(10A=\frac{10^{21}-60}{10^{21}-6}=\frac{10^{21}-6-54}{10^{21}-6}=1-\frac{54}{10^{21}-6}\)

\(10B=\frac{10^{22}-60}{10^{22}-6}=\frac{10^{22}-6-54}{10^{22}-6}=1-\frac{54}{10^{22}-6}\)

ta có: \(10^{21}-6<10^{22}-6\)

=>\(\frac{54}{10^{21}-6}>\frac{54}{10^{22}-6}\)

=>\(-\frac{54}{10^{21}-6}<-\frac{54}{10^{22}-6}\)

=>\(-\frac{54}{10^{21}-6}+1<-\frac{54}{10^{22}-1}+1\)

=>\(\frac{A}{10}<\frac{B}{10}\)

=>A<B

=-42x100=-4200

=-42x100=-4200

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. 

\(M=896+892+888+...+24\)

\(M=\frac{\left(896+24\right).219}{2}=100740\)

\(N=18.21+21+21.79\)

\(N=21.\left(18+1+79\right)=21.100\)

\(N=2100\)

b,

N=18x21+21+21x79

N=18x21+21x1+21x79

N=(18+1+79)x21

N=98x21

N=2058

\(a.x+\left(x+10\right)+\left(x+14\right)+....+\left(x+162\right)=3434\\ x+\left(x+x+.....+x\right)+\left(10+14+18+.....+162\right)=3434\\ x+x+39+3354=3434\\ x\cdot40=80\\ x=2\)

a.x+(x+10)+(x+14)+....+(x+162)=3434

x+(x+x+.....+x)+(10+14+18+.....+162)=3434

x+x.39+3354=3434

x⋅40=80

x=2

\(\dfrac{21}{32}:\dfrac{1}{16}.\dfrac{4}{9}\)

\(=\dfrac{21.16.4}{32.9}\)

\(=\dfrac{3.7.32.2}{32.3.3}\)

\(=\dfrac{14}{3}\)