Xin lỗi em mới lớp ... 4 thôi hà :( 

\(=>P1=U\left(đm1\right).I\left(đm1\right)=220.0,8=176W=>R1=\dfrac{U\left(đm1\right)^2}{P1}=275\Omega\)

\(=>P2=U\left(đm2\right)I\left(đm2\right)=220.0,5=110W=>R2=\dfrac{U\left(đm2\right)^2}{P2}=440\Omega\)

R1 nt R2

\(=>I1=I2=\dfrac{440}{R1+R2}=\dfrac{440}{440+275}=\dfrac{8}{13}A\)

\(=>I1=\dfrac{8}{13}A< I\left(đm1\right),I2=\dfrac{8}{13}A>I\left(đm2\right)\)

=>đèn 1 sáng yếu hơn bth , đèn 2 sáng hơn bth(có thể bị cháy)

do đèn 2 có thể cháy nên ko mắc nối tiếp 

Ex1

1 usually walks

2 does - usually get

3 rarely washes

4 Do - often visit 

5 isn't often

6 usually walks

7 is occasionally 

8 am hardly

9 are always

10 often washes

Ex2

1 are working

3 are making

3 is having

4 is reading

5 is studying

6 am studying

7 is having

8 aren't studying

9 is having

10 are making

Bài 1: 

a: \(A=\left\{2\right\}\)

b: \(B=\left\{0;4;5\right\}\)

một hôm

Một hôm nhâ

Một hôm nhé 

a a b a b a d a

1.a
2.a
3.b

4.b
5.b
6.a
7.d
8.a
 

a c b d c a d a

hơi nhiều nên làm vài bài thoi còn lại bạn tự làm

16. R1 //(R2 nt R3)

\(=>U123=U1=U23=Im.Rtd=2.\left\{\dfrac{R1\left(R2+R3\right)}{R1+R2+R3}\right\}=15\left(V\right)\)

\(=>I1=\dfrac{U1}{R1}=\dfrac{15}{12}=1,25A\)

\(=>I23=I2=I3=\dfrac{U23}{R2+R3}=\dfrac{15}{15+5}=0,75A\)

17. R1 nt(R2 //R3)

a,\(=>Rtd=R1+\dfrac{R2R3}{R2+R3}=4+\dfrac{6.12}{6+12}=8\left(om\right)\)

b,\(=>I1=I23=\dfrac{U}{Rtd}=\dfrac{24}{8}=3A=>U23=U2=U3=I23.R23=3.4=12V=>I2=\dfrac{U2}{R2}=2A,I3=\dfrac{U3}{R3}=1A\)

19. R3 nt(R1//R2)ntRa

\(=>U12=\dfrac{R1R2}{R1+R2}=\dfrac{24R2}{24+R2}=U-Ua-U3=24-1.0,2-1.3,8=>R2=120\left(om\right)\)

20. R3 nt Ra nt(R1//R2)(làm tương tự)

18. (R1 nt R2)//R3

\(a,=>Rtd=\dfrac{R3\left(R1+R2\right)}{R3+R1+R2}=4\left(om\right)\)

\(b,=>U3=U12=12V=>I3=\dfrac{U3}{R3}=1A,=>I1=I2=\dfrac{U12}{R1+R2}=\dfrac{12}{2+4}=2A\)

c,\(=>U1=I1R1=4V=>U2=I2R2=8V\)

(mấy bài còn lại lm tương tự)

theo mình thì câu trên: dưới mẫu trong căn bỏ n^2 ra làm nhân tử chung xong đặt nhân tử chung của cả mẫu là n^2 . câu dưới thì mình k biết!!

\(\lim\dfrac{-3n+2}{n-\sqrt{4n+n^2}}=\lim\dfrac{\left(-3n+2\right)\left(n+\sqrt{4n+n^2}\right)}{\left(n-\sqrt{4n+n^2}\right)\left(n+\sqrt{4n+n^2}\right)}\)

\(=\lim\dfrac{\left(-3n+2\right)\left(n+\sqrt{4n+n^2}\right)}{-4n}=\lim\dfrac{n\left(-3+\dfrac{2}{n}\right)n\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4n}\)

\(=\lim n\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}\)

Do \(\lim\left(n\right)=+\infty\)

\(\lim\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}=\dfrac{\left(-3+0\right)\left(1+\sqrt{0+1}\right)}{-4}=\dfrac{3}{2}>0\)

\(\Rightarrow\lim n\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}=+\infty\)

\(\lim\left(\sqrt[3]{n^3+9n^2}-n\right)=\lim\dfrac{\left(\sqrt[3]{n^3+9n^2}-n\right)\left(\sqrt[3]{\left(n^3+9n^2\right)^2}+n\sqrt[3]{n^3+9n^2}+n^2\right)}{\sqrt[3]{\left(n^3+9n^2\right)}+n\sqrt[3]{n^3+9n^2}+n^2}\)

\(=\lim\dfrac{9n^2}{\sqrt[3]{\left(n^3+9n^2\right)^2}+n\sqrt[3]{n^3+9n^2}+n^2}\)

\(=\lim\dfrac{9n^2}{n^2\sqrt[3]{\left(1+\dfrac{9}{n}\right)^2}+n^2\sqrt[3]{1+\dfrac{9}{n}}+n^2}\)

\(=\lim\dfrac{9}{\sqrt[3]{\left(1+\dfrac{9}{n}\right)^2}+\sqrt[3]{1+\dfrac{9}{n}}+1}\)

\(=\dfrac{9}{\sqrt[3]{\left(1+0\right)^2}+\sqrt[3]{1+0}+1}=\dfrac{9}{3}=3\)