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30 tháng 8 2022

From : ...

To : ...

Subject : The camera

Hi,

I'm ........ Are you the company called : ....... that product camera? I have a litte issue about your camera : Last week, I went to the store and bought a new camera which is made by your company. I like it very much. But this morning, when I'm going to take a picture of a flower, I realize that it isn't working any more. I checked it again and it's still not working. I didn't do anything like drop it to the ground since I bought it. I think there's some problem inside it. I'm really worry about the camera. What do you think the problem is? Can you come to my house and check the camera? I'm free tomorrow. If you can, please come by. Here's my address : ...........

Thank you!

26 tháng 10 2023
Sir A week ago, I bought a Sony digital camera from your store. I was very excited when I brought it home but was very disappointed when I discovered the product was defective and unusable. I am 28 years old and working at a private company. I hope you will replace it. Regarding the specifics of the order, I purchased this Sony - Cyber-shot DSC-HX80 18.2-Megapixel Digital Camera from your store on April 17 using a Visa credit card from I. The order number is D4582135 and I am attaching a copy of my receipt. The camera problem is duplicated when it turns off suddenly, strange dots appear on the LCD screen and I can't access the photos in the internal memory so I'm sure this is a manufacturing defect. As a solution, and since I really like the functionality of the camera, I hope that you will agree that there can be errors when using a new and fresh one. If not, I would like a full refund. If I do not hear from you on this matter within a week, I would like to submit a written complaint to your higher authority. Thank you for your interest in this matter and I hope to hear from you soon. Best regards Linh

 

28 tháng 8 2022

Với `x \ne 0,x \ne 1` có:

`A=([x\sqrt{x}]/[\sqrt{x}-1]-[x^2]/[x\sqrt{x}])(1/\sqrt{x}-1)^2`

`A=([x\sqrt{x}]/[\sqrt{x}-1]-x/\sqrt{x})([1-\sqrt{x}]/\sqrt{x})^2`

`A=[x^2-x(\sqrt{x}-1)]/[\sqrt{x}(\sqrt{x}-1)].[(\sqrt{x}-1)^2]/x`

`A=[x(x-\sqrt{x}-1)]/\sqrt{x}.[\sqrt{x}-1]/x`

`A=[x-\sqrt{x}-1]/[\sqrt{x}-1]`

26 tháng 8 2022

`a)`Ta có: \(\left\{{}\begin{matrix}AB\perp AC\\HE\perp AC\end{matrix}\right.\) \(\Rightarrow\)`AB////HE`

`b)`Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow\widehat{C}=180^o-90^o-60^o=30^o\)

Xét tam giác AHC, có:

\(\widehat{HAC}=180^o-30^o-90^o=60^o\)

\(\widehat{A}=\widehat{BAH}+\widehat{HAC}\)

\(\Rightarrow\widehat{BAH}=90^o-60^o=30^o\)

Ta có: \(\widehat{BAH}=\widehat{AHE}=30^o\) ( so le trong )

26 tháng 8 2022

Vì \(\widehat{xBA}=\widehat{BAD}\left(=50^o\right)\) mà \(\widehat{xBA}\text{ và }\widehat{BAD}\) là 2 góc so le trong

=> Bx//AD (1)

Vì \(\widehat{DAC}=\widehat{ACy}\left(=30^o\right)\) mà \(\widehat{DAC}\text{ và }\widehat{ACy}\) là 2 góc so le trong

=> AD // Cy (2)

Từ (1) và (2) => Bx // Cy

26 tháng 8 2022

Ta có:

`@` \(\widehat{ABx}=\widehat{DAB}=50^o\)

`=>Bx////AD` ( 2 góc so le trong bằng nhau ) (1)

`@`\(\widehat{ACy}=\widehat{DAC}=30^o\)

`=>Cy////AD` ( 2 góc so le trong bằng nhau ) (2)

\(\left(1\right);\left(2\right)\Rightarrow\)`Bx////Cy`

24 tháng 8 2022

cô mà không giải đc sao tụi nào giải đc đồ đi*ên

31 tháng 8 2022

phonissong

19 tháng 8 2022

số tiền lãi mà mẹ minh được nhận sau 6 tháng là

2062400 - 2000000 = 62400 (đồng)

mỗi tháng số tiền lãi mà mẹ minh nhận được là

62400 : 6 = 10400 (đồng)

% lãi hàng tháng là 

10400 : 2000000 x 100 = 0,52% 

đs....

19 tháng 8 2022

số tiền lãi mà mẹ minh được nhận sau 6 tháng là

2062400 - 2000000 = 62400 (đồng)

mỗi tháng số tiền lãi mà mẹ minh nhận được là

62400 : 6 = 10400 (đồng)

% lãi hàng tháng là 

10400 : 2000000 x 100 = 0,52% 

18 tháng 8 2022

\(Q=-2\sqrt{x-3}+1\le1\)

Dấu ''='' xảy ra khi x = 3 

18 tháng 8 2022

.

15 tháng 8 2022

a + b, A=\(\dfrac{x-3\sqrt{x}+2}{x-4\sqrt{x}+3}\) = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\)

ĐKXĐ: \(\sqrt{x}-3\)\(\Leftrightarrow\sqrt{x}\)\(\ne\)3\(\Leftrightarrow\) x\(\ne\)9

 

15 tháng 8 2022

c, \(\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+1}{\sqrt{x}-3}=1+\dfrac{1}{\sqrt{x}-3}\Rightarrow\sqrt{x}-3\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\sqrt{x}-3\) 1 -1
x 16 4

 

10 tháng 8 2022

My sister and I attended the F5 tour at the National Stadium last night. I could sum up the concert in one word, INCREDIBLE. We found our way up to our seats after having a light meal and stood in a queue at the gate of the stadium for 45 minutes. When the curtain was raised to reveal the F5 band, the entire stadium went absolutely crazy. I was thrilled by every of their performances. There was so much emotion in many of their songs, and the way they performed was so terrific. This was such a wonderful experience, a night that I'll never forget. I'm so grateful to have been able to have that experience.

10 tháng 8 2022

Last week , i went to a charity concert with my sister , it called : FPC ( for poor children ) .  Because the fare will be for poor children , I and my sister invited more friends to join us . It was held at Hoang Mai Stadium in Hanoi , at 6p.m. It is 40 minutes form my house to Hanoi if I go by bus , so I and my sister need to be in at 5:30 p.m to catch the bus . At the concert, there were many spectators involved. Featuring famous singers, the concert was more lively than ever. Both the songs and the lively dances are exciting . Although I don't like music, when I came to a charity concert, I felt very happy that I was able to help poor children get food and clothing.