Bài 3: 

b: Ta có: CD=CN

mà CN=BM

nên CD=BM

Xét ΔABC có 

\(\dfrac{BM}{BA}=\dfrac{CD}{CA}\)

nên MD//BC

Xét tứ giác BMDC có MD//BC

nên BMDC là hình thang

mà \(\widehat{MBC}=\widehat{DCB}\)

nên BMDC là hình thang cân

Bài 1: 

a, \(\)\(\)\(=>R2//\left[R4nt\left(R3//R5\right)\right]\)

\(=>Rtd=\dfrac{R2\left[R4+\dfrac{R3.R5}{R3+R5}\right]}{R2+R4+\dfrac{R3.R5}{R3+R5}}=\dfrac{1.\left[1+\dfrac{1}{1+1}\right]}{1+1+\dfrac{1}{1+1}}=0,6\left(ôm\right)\)

\(=>I=\dfrac{Uab}{Rtd}=\dfrac{10}{0,6}=\dfrac{50}{3}A=I1\)

\(=>Uab=U2345=10V=U2=U345\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{10}{1}=10A\)

\(=>I345=\dfrac{U345}{R345}=\dfrac{10}{1+\dfrac{1.1}{1+1}}=\dfrac{20}{3}A=I4=I35\)

\(=>U35=I35.R35=\dfrac{20}{3}.\dfrac{1.1}{1+1}=\dfrac{10}{3}V=U3=U5\)

\(=>I3=\dfrac{U3}{R3}=\dfrac{\dfrac{10}{3}}{1}=\dfrac{10}{3}A,\)

\(=>I5=\dfrac{U5}{R5}=\dfrac{10}{3}A\)

b, \(I1=0,1A=Im=I2345\)

\(=>Uab=I2345.R2345=0,1.\dfrac{6\left[8+\dfrac{6.12}{6+12}\right]}{6+8+\dfrac{6.12}{6+12}}=0,4V\)

` x xx 1/3 : 2/5 =4/3`

`=> x xx 1/3 = 4/3 xx 2/5`

`=> x xx 1/3 = 8/15`

`=> x= 8/15 :1/3`

`=> x= 8/15 xx 3`

`=> x= 24/15`

`=> x= 8/5`

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}\)

=\(\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{19}{81.100}\)

=\(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Mà \(\dfrac{99}{100}< 1\) nên \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}< 1\)

Bài 3:

a. \(R=R1+R2=15+30=45\Omega\)

b. \(\left\{{}\begin{matrix}I=U:R=9:45=0,2A\\I=I1=I2=0,2A\left(R1ntR2\right)\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}U1=R1.I1=15.0,2=3V\\U2=R2.I2=30.0,2=6V\end{matrix}\right.\)

Bài 4:

\(I1=U1:R1=6:3=2A\)

\(\Rightarrow I=I1=I2=2A\left(R1ntR2\right)\)

\(U=R.I=\left(3+15\right).2=36V\)

\(U2=R2.I2=15.2=30V\)

1,D

2,B

3,B

4,C

1,D

2,A

3,A

4,D

5,C

Xét ΔOHA vuông tại H và ΔOKA vuông tại K có

OA chung

OH=OK

Do đó: ΔOHA=ΔOKA

Suy ra: AH=AK

hay AB=AC