Hobby is one of the most important things we should have in our lives. It is something that you enjoy doing, something that brings relief of the daily grind and allows you to relax. Of all hobbies in the world, I like music the most. For me, music is the best hobby ever. You should look around and you can see that music is everywhere.

I enjoy music and I listen it every single day. There are a lot of different styles of music for you to choose and enjoy. If I am in good mood, I prefer to listen some energy rock music that keep me happy and full of power. If I am tired, I chose some calm, relaxing songs. Music can speak, through music people can express feelings and emotions. When you hear something that looks like you, that you understand it easily, you start to love it. There are several different ways to enjoy music as your hobby. When you discover an artist or musician that you satisfies, it’s clear that you want to collect all of their works. Not only listening to music, a music collection is a great hobby to have, too. 
Music is a hobby that has no boundaries. It makes the world smaller. Without music, life would be a mistake.

      No one can deny the fact that music plays an important part in our life .It comes to our soul with everlasting songs and unforgetable melodies.

     In fact, music has many benefits. Listening to music is the best way to relax. Scientists have proved that music is good for young children, so the mother should listen to music right from the pregnancy to help develop her baby's intelligence.

     As you know, different people have different tastes in music. Personally, I like all types of music, but my favourite music is piano music because the melodies are gantle and deep. Listening to piano music makes me fell better after hard-working hours. I not only listen to music but I also play the piano and sing a lot, especialy English songs, which helps me improve my English.

     In short, we hardly live without music because music makes uor lives more interesting and relaxing

CHÚC BẠN HỌC TỐT NHÉ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(3x^2-2x.\left(5+1,5x\right)+10\)

\(=3x^2-2x.5-2x.1,5x+10\)

\(=3x^2-3x^2-10x+10\)

\(=10-10x\)

\(=10.\left(1-x\right)\)

cam on ban nha

15 A

16 D

17 B

18 D

19 D

20 A

21 B

22 D

23 A

24 D

24 B

26 C

27 B

28 A

29 D

30 A

15 A

16 D

17 B

18 D

19 D

20 A

21 B

22 D

23 A

24 D

24 B

26 C

27 B

28 A

29 D

30 A

Điện trở của đèn:

\(R_Đ=\dfrac{U_{đm}^2}{P_{đm}}=\dfrac{12^2}{12}=12\left(\Omega\right)\)

Điện trở tương đương nhánh song song:

\(R_{Đb}=\dfrac{R_Đ.R_b}{R_Đ+R_b}=\dfrac{12.36}{12+36}=9\left(\Omega\right)\)

Điện trở tương đương của đoạn mạch:

\(R_{tđ}=R_1+R_{Đb}=6+9=15\left(\Omega\right)\)

Cường độ dòng điện qua mạch chính:

\(I=\dfrac{U}{R_{tđ}}=\dfrac{20}{15}=\dfrac{4}{3}\left(A\right)\)

\(I=I_1=I_{Đb}=\dfrac{4}{3}\left(A\right)\)

Hiệu điện thế ở nhánh song song:

\(U_{Đb}=I_{Đb}.R_{Đb}=\dfrac{4}{3}.9=12\left(V\right)\)

\(U_{Đb}=U_Đ=U_b=12\left(V\right)\)

Đèn sáng bình thường do \(U_Đ=D_{đm}\left(12=12\right)\)

b) Gọi x là là giá trị biến trở để công suất trên biến trở là lớn nhất

Điện trở tương đương:

\(R=R_1+\dfrac{R_Đ.x}{R_Đ+x}=6+\dfrac{12x}{12+x}=\dfrac{6\left(12+x\right)}{12+x}+\dfrac{12x}{12+x}=\dfrac{72+6x+12x}{12+x}=\dfrac{72+18x}{12+x}\left(\Omega\right)\)Cường độ dòng điện của đoạn mạch:

\(I=\dfrac{U}{R}=\dfrac{20}{\dfrac{72+18x}{12+x}}=\dfrac{20\left(12+x\right)}{72+18x}\left(A\right)\)

\(I=I_1=I_{Đb}=\dfrac{20\left(12+x\right)}{72+18x}\left(A\right)\)

Hiệu điện thế nhánh song song:

\(U_{Đb}=I_{Đb}.R_{Đb}=\dfrac{20\left(12+x\right)}{72+18x}.\dfrac{12x}{12+x}=\dfrac{240x}{72+18x}\left(V\right)\)

\(U_{Đb}=U_Đ=U_b=\dfrac{240x}{72+18x}\left(V\right)\)

Công suất trên \(R_b:\)

\(P_b=\dfrac{U_b^2}{R_b}=\dfrac{\left(240x\right)^2}{\left(72+18x\right)^2.x}=\dfrac{240^2.x}{\left[72^2+2.72.18x+\left(18x\right)^2\right]}\\ =\dfrac{240^2}{\dfrac{72^2}{x}+2.72.18+18^2x}\left(W\right)\)

Để \(P_{b-max}\) thì \(\dfrac{72^2}{x}+2.72.18x+18^2x\) đạt GTNN

\(\rightarrow\dfrac{72^2}{x}+18^2x\) đạt GTNN

Áp dụng bất đẳng thức Cauchy schwarz, ta có:

\(\dfrac{72^2}{x}+18^2x\ge2\sqrt{\dfrac{72^2}{x}.18^2x}=2\sqrt{72^2.18^2}=2.72.18=2592\)

Dấu \("="\) xảy ra khi và chỉ khi 

\(\dfrac{72^2}{x}=18^2x\\ \rightarrow72^2=18^2x^2\\ \rightarrow x^2=16\\ \rightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

Công suất cực đại là \(\dfrac{240^2}{2592+2.72.18}=\dfrac{100}{9}\left(W\right)\)

Vậy công suất cực đại trên Rb \(R_{b-max}=\dfrac{100}{9}W\) khi \(R_b=4\Omega\)

Tui ngồi miệt mài 1 tiếng làm cho á, 5' kiểm tra lại bài nữa nhưng bài này đáng giá quá <3

Does she like nuts?

=> Yes ,she does.

What does she like ?

=> She likes Sweets.

Does he like balloons? 

=> No, he doesn't.

Does he like sweets?

=>No,he doesn't.

Câu 7: C

Câu 8: B

Alo. Cho tớ trả lời nhé :

5a

6b

7c ( Tớ lưu ý chút, trong PowerPoint 2007 và các phiên bản trở xuống, câu b sẽ là câu trả lời trong câu hỏi này, các phiên bản trở lên như PowerPoint 2010, sẽ chọn câu c)

8b

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1}{2}\\x_1x_2=-2\end{matrix}\right.\)

\(A=3-x_1^2-x_2^2\\ =3-\left(x_1^2+x_2^2\right)\\ =3-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =3-\left[\left(-\dfrac{1}{2}\right)^2-2.\left(-2\right)\right]\\ =3-\left(\dfrac{1}{4}+4\right)\\ =3-\dfrac{17}{4}\\ =-\dfrac{5}{4}\)

\(B=\left(x_1-x_2\right)^2\\ =x_1^2+x_2^2-2x_1x_2\\ =\left(x_1+x_2\right)^2-4x_1x_2\\ =\left(\dfrac{1}{2}\right)^2-4.\left(-2\right)\\ =\dfrac{1}{4}+8\\ =\dfrac{33}{4}\)

\(D=\left(1+x_1\right)\left(2-x_1\right)+\left(1+x_2\right)\left(2-x_2\right)\\ =2+x_1-x_1^2+2+x_2-x_2^2\\ =4+\left(x_1+x_2\right)-\left(x_1^2+x_2^2\right)\\ =4+\dfrac{1}{2}-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\dfrac{9}{2}-\left[\left(\dfrac{1}{2}\right)^2-2.\left(-2\right)\right]\\ =\dfrac{9}{2}-\dfrac{17}{4}\\ =\dfrac{1}{4}\)

Chọn A