A= (2+2²)+(2³+2⁴)+...+(2⁵⁹+2⁶⁰)

A=2.(1+2)+2³.(1+2)+...+2⁵⁹.(1+2)

A=2.3+2³.3+...+2⁵⁹.3

A=3.(2+2³+...+2⁵⁹)

Vì 3 chia hết cho 3 => 3.(2+2³+...+2⁵⁹)  chia hết cho 3

=>A  chia hết cho 3

A = 2 + 2² + 2³ + 2⁴ + ... + 2⁶⁰

=> A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ... + ( 2⁵⁹ + 2⁶⁰ )

=> A = 2( 1 + 2 ) + 2²(1 + 2 ) + ... + 2⁵⁹( 1 + 2 )

=> A = 2 . 3 + 2² . 3 + ... + 2⁵⁹ . 3

=> A = ( 2 + 2² + ²⁵⁹ ) . 3 chia hết cho 3

Vậy A chia hết cho A

The better a child is brought up, the better he behaves 

The faster you drive, the more likely you are to have an accident

The less you work, the lower marks you’ll have in your exam

The harder you study, the easier you’ll find exams

The longer I waited for him, the more worried I got

The more sport I practised, the fitter I got

The more I explained it to him, the less he understood

The more clothes I put on, the colder I am

The longer I know him, the less I like him

The cleverer the students are, the easier it is to teach them

The sooner you come, the better

The more attention you pay to him, the worse

The more pleasant atmosphere you work in, the less stressed you’ll be

The more careful you are when driving, the fewer accidents you’ll have

The more money he earns, the more his wife complains

The more exercises you do, the better you’ll learn the subject

The longer I listened to his story, the more surprise I was

The farther from your school you live, the earlier you'll have to get

Đâu bạn... 

6. We haven't seen a movie for 2 months

7. She hasn't gone out with her parents for 4 months

8. An exam has been taken since 10 years ago

9. My friend has studied Chinese for 2 months.

Thì HTHT nhé

\(x\in\left(\dfrac{\pi}{4};\dfrac{3\pi}{4}\right)\Rightarrow2x\in\left(\dfrac{\pi}{2};\dfrac{3\pi}{2}\right)\)

a.

D E thuộc Ox \(\Rightarrow\) tọa độ E có dạng \(E\left(x;0\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{OE}=\left(x;0\right)\\\overrightarrow{OM}=\left(4;1\right)\end{matrix}\right.\)

Tam giác OEM cân tại O \(\Rightarrow OE=OM\)

\(\Rightarrow\sqrt{x^2+0^2}=\sqrt{4^2+1^2}\Rightarrow x^2=17\)

\(\Rightarrow x=\pm\sqrt{17}\Rightarrow\left[{}\begin{matrix}E\left(\sqrt{17};0\right)\\E\left(-\sqrt{17};0\right)\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}\overrightarrow{MA}=\left(a-4;-1\right)\\\overrightarrow{MB}=\left(-4;b-1\right)\end{matrix}\right.\)

Tam giác ABM vuông tại M \(\Rightarrow\overrightarrow{MA}.\overrightarrow{MB}=0\)

\(\Rightarrow-4\left(a-4\right)-1\left(b-1\right)=0\)

\(\Leftrightarrow4a+b-17=0\Rightarrow b=17-4a\)

Lại có \(S_{ABM}=\dfrac{1}{2}MA.MB=\dfrac{1}{2}\sqrt{\left(a-4\right)^2+1}.\sqrt{\left(b-1\right)^2+16}\)

\(=\dfrac{1}{2}\sqrt{\left(a-4\right)^2+1}.\sqrt{\left(16-4a\right)^2+16}=\dfrac{1}{2}\sqrt{\left(a-4\right)^2+1}.\sqrt{16\left[\left(a-4\right)^2+1\right]}\)

\(=2\left[\left(a-4\right)^2+1\right]\ge2\)

Dấu "=" xảy ra khi \(a-4=0\Rightarrow a=4\Rightarrow b=1\)