Câu 20: C

Câu 21: B

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2) \(\dfrac{\left(1+\sqrt{a}\right)^2-\left(2-\sqrt{a}\right)^2}{1-2\sqrt{a}}:\dfrac{\sqrt{a}}{3}\left(a>0,a\ne\dfrac{1}{4}\right)\)

\(=\dfrac{\left(1+\sqrt{a}-2+\sqrt{a}\right)\left(1+\sqrt{a}+2-\sqrt{a}\right)}{1-2\sqrt{a}}.\dfrac{3}{\sqrt{a}}\)

\(=\dfrac{3.\left(2\sqrt{a}-1\right)}{1-2\sqrt{a}}.\dfrac{3}{\sqrt{a}}=-\dfrac{9}{\sqrt{a}}\)

5) \(\left(5-\dfrac{a+3\sqrt{a}}{\sqrt{a}+3}\right)\left(2-\dfrac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)\left(a\ge0\right)\)

\(=\left(5-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}\right)\left(2-\dfrac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)

\(=\left(5-\sqrt{a}\right)\left(2-\sqrt{a}\right)=10-7\sqrt{a}+a\)

6) \(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)\left(a,b\ge0,a\ne9,b\ne25\right)\)

\(=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2+\dfrac{\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)

\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\)

3) Ta có: \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)

\(=\sqrt{a}+2-\sqrt{a}-2\)

=0

\(S=\dfrac{1}{2^2}+\dfrac{1}{\left(2.2\right)^2}+\dfrac{1}{\left(2.3\right)^2}+...+\dfrac{1}{\left(2.10\right)^2}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{2^2.2^2}+\dfrac{1}{2^2.3^2}+...+\dfrac{1}{2^2.10^2}\)

\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\right)\)

\(< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)\)

\(=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{4}\left(2-\dfrac{1}{10}\right)< \dfrac{1}{4}.2=\dfrac{1}{2}\) (đpcm)

Bạn tự vẽ sơ đồ mạch điện và tự làm tóm tắt nhé!
Bài 1:

b. \(U=IR=I\left(R1+R2\right)=0,4\left(15+2\right)=6,8\left(V\right)\)

c. \(I=I1=I2=\dfrac{U'}{R}=\dfrac{60}{15+2}=\dfrac{60}{17}\simeq3,5\left(A\right)\left(R1ntR2\right)\)

Bài 2:

\(5400kJ=1500\left(Wh\right)\)

a. \(A=Pt\Rightarrow P=\dfrac{A}{t}=\dfrac{1500}{1}=1500\left(W\right)\)

b. \(P=UI\Rightarrow I=\dfrac{P}{U}=\dfrac{1500}{220}=\dfrac{75}{11}\simeq6,82\left(A\right)\)

Bài 3:

a. \(R=p\dfrac{l}{S}=1,1.10^{-6}\dfrac{3}{0,05.10^{-6}}=66\left(\Omega\right)\)

b. \(P=UI=U\left(\dfrac{U}{R}\right)=220.\left(\dfrac{220}{66}\right)=733,33\left(W\right)\)

c. \(A=Pt=733,33.\left(\dfrac{30}{60}\right)=366,665\left(Wh\right)=0,366665\left(kWh\right)=1319994\left(J\right)\)

Bài 4:

a. \(S=\pi\dfrac{d^2}{4}=\pi\dfrac{1^2}{4}=0,785\left(mm^2\right)\)

\(\Rightarrow R=p\dfrac{l}{S}=5,5.10^{-8}\dfrac{10}{0,785.10^{-6}}=\dfrac{110}{157}\simeq0,7\left(\Omega\right)\)

b. \(A=Pt=UIt=U\left(\dfrac{U}{R}\right)t=70\left(\dfrac{70}{0,7}\right).\dfrac{1}{3}=2333,33\left(Wh\right)=2,33333\left(kWh\right)\simeq8400000\left(J\right)\)

Câu 6:

\(U=I.R=10.1,5=15\left(V\right)\)

Câu 7:

\(P=U.I\Rightarrow U=\dfrac{P}{I}=\dfrac{10}{0,5}=20\left(V\right)\)

Câu 8:

\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{10.30}{10+30}=7,5\left(\Omega\right)\)

1 C

2 C

3 C

4 C

5 A

6 B

7 A

11 

12 B

13 A

14 C

15D

4. B

5. A

6. C

7. C

8. B

9. B

10. A

11. C

12. 

13. A

14. 

15. C