b: Xét ΔABH vuông tại H có 

\(AB^2=AH^2+HB^2\)

hay AH=12(cm)

Xét ΔAHB vuông tại H có 

\(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AH}{AB}=\dfrac{12}{13}\)

\(\cos\widehat{B}=\sin\widehat{C}=\dfrac{5}{13}\)

\(\tan\widehat{B}=\cot\widehat{C}=\dfrac{12}{5}\)

\(\cot\widehat{B}=\tan\widehat{C}=\dfrac{5}{12}\)

a. \(I=I1=I23=U:R=6:\left(1,8+\dfrac{3\cdot2}{3+2}\right)=2A\left(R1ntR23\right)\)

\(\rightarrow U1=I1\cdot R1=2\cdot1,8=3,6V\)

\(\rightarrow U23=U2=U3=I23\cdot R23=2\cdot\left(\dfrac{3\cdot2}{3+2}\right)=2,4V\left(R1//R2\right)\)

\(\Rightarrow\left\{{}\begin{matrix}I2=U2:R2=2,4:3=0,8A\\I3=U3:R3=2,4:2=1,2A\end{matrix}\right.\)

b. \(\left\{{}\begin{matrix}Q_{toa1}=A_1=U_1\cdot I_1\cdot t=3,6\cdot2\cdot1\cdot60=432\left(J\right)\\Q_{toa2}=A_2=U_2\cdot I_2\cdot t=2,4\cdot0,8\cdot1\cdot60=115,2\left(J\right)\end{matrix}\right.\)

Câu 36: -6(x+7)=96

=>x+7=-16

=>x=-16-7

=>x=-23

=>Chọn D

Câu 35: B

Câu 34: -5-x=-11

=>x+5=11

=>x=11-5

=>x=6

=>Chọn A

câu 33: B

câu 32: A

Câu 31: C

Câu 30: A

Câu 29: B

Câu 28: A

Câu 22: A

Câu 23: A

Câu 24: B

Câu 25: A

Câu 26: C

Câu 27: D

Câu 10: D

Câu 11: B

Câu 12: A

Câu 13:A

Câu 15: D

Câu 16: C

Câu 17: C

câu 18: B

Câu 19: A

Câu 20: D

Câu 21: B

a: Xét ΔBHA vuông tại H và ΔBAC vuông tại A có

\(\hat{HBA}\) chung

Do đó; ΔBHA~ΔBAC

=>\(\frac{BH}{BA}=\frac{BA}{BC}\)

=>\(BH\cdot BC=BA^2\)

b: Xét ΔHAC vuông tại H và ΔHDB vuông tại H có

\(\hat{HAC}=\hat{HDB}\) (hai góc so le trong, AC//DB)

Do đó: ΔHAC~ΔHDB

=>\(\frac{HA}{HD}=\frac{HC}{HB}\)

=>\(HA\cdot HB=HD\cdot HC\)

c:

BD//AC

AB⊥ AC

Do đó: BD⊥BA

XétΔABD vuông tại B và ΔCAB vuông tại A có

\(\hat{BAD}=\hat{ACB}\left(=90^0-\hat{ABC}\right)\)

Do đó: ΔABD~ΔCAB

d: ΔHAC~ΔHDB

=>\(\frac{HA}{HD}=\frac{AC}{DB}=\frac{AN}{DM}\)

Xét ΔHAN và ΔHDM có

\(\frac{HA}{HD}=\frac{AN}{DM}\)

\(\hat{HAN}=\hat{HDM}\) (hai góc so le trong, AN//DM)

Do đó: ΔHAN~ΔHDM

=>\(\hat{AHN}=\hat{DHM}\)

mà \(\hat{AHN}+\hat{NHD}=180^0\) (hai góc kề bù)

nên \(\hat{DHM}+\hat{NHD}=180^0\)

=>N,H,M thẳng hàng

Tất cả k dưới đây là \(k\in Z\)

1.

ĐKXĐ: \(1-2cosx\ne0\Rightarrow cosx\ne\dfrac{1}{2}\)

\(\Rightarrow x\ne\pm\dfrac{\pi}{3}+k2\pi\)

2.

\(cos2x-1=0\Rightarrow cos2x=1\)

\(\Rightarrow2x=k2\pi\)

\(\Rightarrow x=k\pi\)

b.

\(\sqrt{3}cotx-3=0\Rightarrow cotx=\sqrt{3}\)

\(\Rightarrow x=\dfrac{\pi}{6}+k\pi\)

c.

\(2sin^22x+sin2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{2}+k2\pi\\2x=\dfrac{\pi}{6}+k2\pi\\2x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

3.

Theo điều kiện có nghiệm của pt lượng giác bậc nhất, pt đã cho có nghiệm khi:

\(\sqrt{3}^2+2^2\ge m^2\)

\(\Rightarrow m^2\le7\)

\(\Rightarrow-\sqrt{7}\le m\le\sqrt{7}\)

a: Đơn chất: \(H_2;N_2;Cl_2;O_2\)

a: Đơn chất: 

đề đâu ????

bạn thấy đề chưa vậy

2 she had got the price of the contest, she would be happy now

3 I had gone to the concert last night ,I would be tired

4 John had finished all his homework, he could see a movie now

5 we had listened to my mother, we wouldn't be in trouble right now

Ex5

1 you arrive at the offcie earlier than I do, please turn on the air-conditioner

2 Were it not snowy, the children would go to school

3 he not died so young, he would be a famous musician

4 Were I you, I would take care of my car

5 not very bad-tempered, his wife would give him soon are marrage

6 Had she not got married at such an early age, he would be at university