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30 tháng 5 2023

Pattern 2

0 Don't forget to write a sick note when you're sick

1 Don't forget to brush your teeth regularly

2 Don't forget to wash vegetables carefully

3 Remember to take morning exercises regulary

4 Don't forget not to eat too much candy

5 Don't forget to iron and wash your own clothes

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Pattern 3

0 I had a stomachache last night

1 My tooth hurts

2 Her stomach hurt yesterday

3 His head hurt last night

4 Minh's tooth hurts

5 Lan had a headache yesterday

31 tháng 5 2023

 Do \(CA=CB=a\) nên \(BE.BC+AC.AK=a\left(AK+BE\right)\) 

 Ta chứng minh \(AK+BE\) không đổi. Thật vậy, gọi P là giao điểm của KE và AB. Quan sát thấy E là trực tâm tam giác ABK \(\Rightarrow KP\perp AP\) tại P. Lại có \(\widehat{KAP}=45^o\) nên suy ra \(\widehat{AKP}=45^o\). Từ đó suy ta tam giác CEK cân tại C hay \(CE=CK\)

 Từ đó \(AK+BE=AC+CK+BC-CE=2a\). Vậy \(BE.BC+AC.AK=2a^2\) không đổi (đpcm)

30 tháng 5 2023

VI )

1) This shirt is too small for me to wear.
=> This shirt isn’t big enough for me to wear
2) That suitcase isn’t big enough to hold all my clothes.
=> That suitcase is so small that it can't hold all my clothes
3) I get home too late to watch the show.
=> I don’t get home early enough to watch the show
4) This computer isn’t fast enough for us to handle the job.
=> It is such a slow computer that we can't handle the job
5) I was too sick to do my homework.
=> I was so sick that I can't do my homework
6) I am too poor to buy a car.
=> I am not poor enough to buy a car
7) The weather was too bad for her to go for a picnic.
=> The weather was so bad that she couldn't go for a picnic
8) Tom is too young to see the horror film.
=> Tom is so young that he can't see the horror film
9) He drank so much coffee that he could get to sleep.
=> He drank such a lot of coffee that he could get to sleep
10) The door was too heavy for the child to push on.
=> It was such a heavy door that the child couldn't push on.

30 tháng 5 2023

VII

She is a good cook

He is a slow cyclist

My uncle is a good English teacher

Many students were involved in the campaign against air pollution 

He is a careful typist

He is a good guitar player

He was severely punished from the master

His father takes up fishing

Rita is proud of her family tradition

These children can run quickly

30 tháng 5 2023

Ta có : \(S_{ABC}=\dfrac{AH.BC}{2}\)

Kẻ đường cao từ B xuống AC tại E do đó :

\(S_{ABC}=\dfrac{BE.AC}{2}\)

mà \(BE< AB\) ( AB là cạnh huyền trong tam giác ABE )

Do đó :

\(\dfrac{AB.AC}{2}\ge\dfrac{BE.AC}{2}=\dfrac{AH.BC}{2}\)

\(\Rightarrow AB.AC\ge AH.BC\left(đpcm\right)\)

Dấu bằng xảy ra khi và chỉ khi : BE trùng với AB

\(\Leftrightarrow\Delta ABC\) vuông tại A .

 

30 tháng 5 2023

ĐKXĐ : \(x\ge\dfrac{1}{2}\)

Đặt \(\sqrt{2x-1}=a;\sqrt{8x+1}=b\left(a;b\ge0\right)\)

=> \(a^2=2x-1;b^2=8x+1\Rightarrow\dfrac{a^2+b^2}{10}=x\)

Lại có \((13x+1).\sqrt{2x-1}=(7x-1).\sqrt{8x+1}-4\)

\(\Leftrightarrow-\left(\sqrt{2x-1}\right)^3+15x.\sqrt{2x-1}=-\left(\sqrt{8x+1}\right)^3+15x.\sqrt{8x+1}-4\)

\(\Leftrightarrow-a^3+15ax=-b^3+15bx-4\)

\(\Leftrightarrow a^3-b^3-\dfrac{3}{2}.\left(a-b\right).\left(a^2+b^2\right)=4\)

\(\Leftrightarrow-\left(a-b\right)^3=8\)

\(\Leftrightarrow a=b-2\)

Thay vào ta được : \(\sqrt{2x-1}=\sqrt{8x+1}-2\)

\(\Leftrightarrow3x+3=2\sqrt{8x+1}\) 

\(\Leftrightarrow\left\{{}\begin{matrix}9x^2-14x+5=0\\x\ge-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{9}\end{matrix}\right.\)(tm ĐKXĐ)

 

30 tháng 5 2023

474+7+7=2(474+7+7)2=8278+27+142=727+17+27+1+142=(71)2(7+1)2+142=717+1+142=7171+142=1422=2(72)2=72

AH
Akai Haruma
Giáo viên
30 tháng 5 2023

Lời giải:
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\sqrt{\frac{8-2\sqrt{7}}{2}}-\sqrt{\frac{8+2\sqrt{7}}{2}}=\sqrt{\frac{(\sqrt{7}-1)^2}{2}}-\sqrt{\frac{(\sqrt{7}+1)^2}{2}}\)

\(=\frac{|\sqrt{7}-1|}{\sqrt{2}}-\frac{|\sqrt{7}+1|}{\sqrt{2}}=\frac{\sqrt{7}-1-(\sqrt{7}+1)}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

30 tháng 5 2023

P nguyên <=> \(\dfrac{x-5}{\sqrt{x}+1}\) nguyên

<=> \(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{4}{\sqrt{x}+1}\) nguyên

<=> \(\sqrt{x}-1-\dfrac{4}{\sqrt{x}+1}\) nguyên

=> \(\sqrt{x}+1\inƯ\left(4\right)=\left\{1;2;4\right\}\)  (vì \(\sqrt{x}+1>0\forall x\in N\))

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