Bài 2:

a: \(\left(-2,5\right)\cdot0,65-1,5\cdot0,3+\left(-0,35\right)\cdot1,5\)

\(=-2,5\cdot0,65-1,5\cdot\left(0,3+0,35\right)\)

\(=-2,5\cdot0,65-1,5\cdot0,65\)

\(=0,65\left(-2,5-1,5\right)=0,65\cdot\left(-4\right)=-2,6\)

b: \(2,51\cdot\left(-0,3\right)+\left(-0,73\right)-0,3\cdot\left(-1,51\right)-2,27\)

\(=-0,3\left(2,51-1,51\right)-0,73-2,27\)

=-0,3-3

=-3,3

c: \(\left\lbrack\left(-0,7\right):0,07+2,5\right\rbrack\cdot\left(1,3\cdot2,5+1,3\cdot7,5\right)\)

\(=\left\lbrack-10+2,5\right\rbrack\cdot1,3\left(7,5+2,5\right)\)

\(=-7,5\cdot1,3\cdot10=-75\cdot1,3=-97,5\)

d: \(17,58\cdot43+57\cdot17,58\)

\(=17,58\left(43+57\right)\)

\(=17,58\cdot100=1758\)

Bài 1:

a: 32-(-1,6)=32+1,6=33,6

b: (-0,5)-1,23=-0,5-1,23=-1,73

c: (-4,5)+3,6+4,5+(-3,6)

=-4,5+4,5+3,6-3,6

=0

d: 2,1+4,2+(-7,9)+(-2,1)+7,9

=2,1-2,1+7,9-7,9+4,2

=0+0+4,2

=4,2

Haizzzzz

câu 1           C

câu 2           C

câu 3            A

câu 4            D

câu 5                      

1:D

2:C

3:A

4:D

5:A

6:B

7:D

8:B

9:A

10:B

11:C

12:D

13:A

14:B

15:D

16:D

17:B

18:D

19:D

20:C

1: Xét ΔABD vuông tại A và ΔDAC vuông tại D có

góc ABD=góc DAC

=>ΔABD đồng dạng với ΔDAC

2: ΔABD đồng dạng với ΔDAC

=>BD/AC=AB/DA=AD/DC

=>AD/16=BD/AC=18/DA

=>AD^2=16*18=288

=>AD=12căn 2(cm)

AC=căn AD^2+DC^2=4căn 82(cm)

quê hương là nơi sinh ra, lớn lên, là nơi ghi dấu những kỉ niệm tuổi thơ.

ăn cắp trên mạng hả thu trang

Bài 1:

a. \(R=p\dfrac{l}{S}=1,10.10^{-6}\dfrac{30}{0,3\cdot10^{-6}}=110\Omega\)

b. \(I=U:R=220:110=2A\)

Bài 2:

a. \(R=R1+R2=30+50=80\Omega\)

b. \(I=I1=I2=0,25A\left(R1ntR2\right)\)

\(\left\{{}\begin{matrix}U1=I1\cdot R1=0,25\cdot30=7,5V\\U2=I2\cdot R2=0,25\cdot50=12,5V\\U=IR=0,25\cdot80=20V\end{matrix}\right.\)

Câu 1.

a)\(R=\rho\cdot\dfrac{l}{S}=1,1\cdot10^{-6}\cdot\dfrac{30}{0,3\cdot10^{-6}}=110\Omega\)

b)\(I=\dfrac{U}{R}=\dfrac{220}{110}=2A\)

Câu 2.

a)\(R_{AB}=R_1+R_2=30+50=80\Omega\)

b)\(I_1=I_2=I_A=0,25A\)

   \(U_1=R_1\cdot I_1=30\cdot0,25=7,5V\)

   \(U_2=R_2\cdot I_2=50\cdot0,25=12,5V\)

   \(U_{AB}=U_1+U_2=7,5+12,5=20V\)

Câu 3.

a)\(R_{AB}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{600\cdot900}{600+900}=360\Omega\)

b)\(U_1=U_2=U_m=220V\)

   \(I_1=\dfrac{U_1}{R_1}=\dfrac{220}{600}=\dfrac{11}{30}A\)

   \(I_2=\dfrac{U_2}{R_2}=\dfrac{220}{900}=\dfrac{11}{45}A\)

   \(I_m=I_1+I_2=\dfrac{11}{30}+\dfrac{11}{45}=\dfrac{11}{18}A\)

1. In spite of being a poor student, he studied very well

2. Despite the bad weather, she went to school on time

3. Despite having a physical handicap, she has become a successful woman

4. In spite of having not finished the paper, he went to sleep

5. Despite having a lot of noise in the city, I prefer living there

1. In spite of being a poor student, he studied very well

2. Despite the fact that the weather was bad, she went to school on time.

\(CaCO_3\underrightarrow{1}CO_2\underrightarrow{2}Na_2CO_3\underrightarrow{3}NaOH\underrightarrow{4}Na_2SO_4\)

(1) \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

(2) \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

(3) \(Na_2CO_3+Ca\left(OH\right)_2\rightarrow2NaOH+CaCO_3\)

(4) \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

 Chúc bạn học tốt

\(\left(1\right)CaCO_2\rightarrow^{t^o}CaO+CO_2\uparrow\\ \left(2\right)CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ \left(3\right)Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3\downarrow+2NaOH\\ \left(4\right)2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Câu 24: A

Câu 25: D

Câu 26: B

Câu 27: B

Câu 28: A

a

d

b

a

c

sin? =2/5

đề thiếu nhé