không thể tìm được bạn ạ :D

U đâu mà tìm

Program    tinh;

Uses          crt;

Var 

a,b,tong,hieu:Integer;

Begin

If a>=b then ( neu a lon hon hoac bang b thi thong bao );

Write('nhap so a la');readln(a)

Write('nhap so b la');readln(b)

tong:=a+b

hieu:a-b

Writeln('tong hai so la',tong);

Writeln('hieu hai so la'hieu);

Readln;

End.

Vì \(AB//CD\) nên \(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=180^0\\\widehat{A}+\widehat{D}=180^0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\left(180^0+40^0\right):2=110^0\\3\widehat{D}=180^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{C}=180^0-110^0=70^0\\\widehat{D}=60^0\end{matrix}\right.\Rightarrow\widehat{A}=120^0\)

\(\widehat{B}=110^0\)

\(\widehat{C}=70^0\)

\(\widehat{A}=120^0\)

\(\widehat{D}=60^0\)

Uhm, sơ đồ mạch điện đâu bạn nhỉ?

a. \(R=R1+R2=5+15=20\left(\Omega\right)\)

\(I=I1=I2=\dfrac{U}{R}=\dfrac{3}{20}=0,15\left(A\right)\left(R2ntR2\right)\)

b. \(\left\{{}\begin{matrix}U1=I1.R1=0,15.5=0,75\left(V\right)\\U2=I2.R2=0,15.15=2,25\left(V\right)\end{matrix}\right.\)

Câu 5:

ABCD là hình bình hành

=>vecto AB=vecto DC

=>\(\left\{{}\begin{matrix}4-x=2-0=2\\-1-y=1+3=4\end{matrix}\right.\Leftrightarrow D\left(2;-5\right)\)

Câu 6:

vecto c=k*vecto a+m*vecto b

=>\(\left\{{}\begin{matrix}-1=2k+3m\\7=-3k+m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}k=-2\\m=1\end{matrix}\right.\)

=>k+m=-1

Câu 7: B

Câu 8: C

\(\dfrac{l1}{l2}=\dfrac{R1}{R2}\Leftrightarrow\dfrac{1}{4}=\dfrac{20}{R2}=>R2=80\Omega\)

Chọn A

Câu 9: B

Câu 10: A

Câu 11: C

Câu 12: C

Câu 13: B