1. In spite of being a poor student, he studied very well

2. Despite the bad weather, she went to school on time

3. Despite having a physical handicap, she has become a successful woman

4. In spite of having not finished the paper, he went to sleep

5. Despite having a lot of noise in the city, I prefer living there

1. In spite of being a poor student, he studied very well

2. Despite the fact that the weather was bad, she went to school on time.

1. Because of studying hard, I passed the exam

2. Because of Hoa's richness, she could buy that house

3. Because of bad grades, she failed the University entrance exam

4. Because of the accident, I was late

5. Because of the terrible traffic, we didn't arrive until 6 o'clock

Đây là đề thi mà? Phải tự làm chứ!

Em cần hỗ trợ bài nào?

ta có các số là 11 ,12,15,22,24,33,36,44,48,55,66,77,88,99

vậy số số lập được là 14

Bạn tách ra đi nhé! Chứ Sử mà như lày thì chết :v

Số nguyên tố nha bn!

tổng trên là số nguyên tố

tk nha

số nguyên tố

A=|x - 2009| + |x - 2010| + |x - 2011| 

*TH1: Xét x ≤ 2009 ; khi đó 

. A = 2009 - x + 2010 - x + 2011 -x 

. A = 6030 - 3x 

có x ≤ 2009 --> -x ≥ -2009 --> -3x ≥ -6027 --> 6030 - 3x ≥ 3 

Dấu " = " <=> x = 2009 

--> Amin = 3 <=> x = 2009 

*TH2 : Xét 2009 < x ≤ 2010 ; ta có 

. A = x - 2009 + 2010 - x + 2011 - x 

. A = 2012 - x 

có x ≤ 2010 --> -x ≥ -2010 --> 2012 - x ≥ 2 

--> Amin = 2 <=> x = 2010 

*TH3 : Xét 2010 < x < 2011 ; ta có : 

. A = x - 2009 + x - 2010 + 2011 - x 

. A = x - 8 > 2010 - 8 = 2002 --> không có min 

*TH4 : Xét x ≥ 2011 ; ta có : 

. A = x - 2009 + x - 2010 + x - 2011 

. A = 3x - 6030 ≥ 3.1011 - 6030 = 3 

Dấu " = " <=> xảy ra <=> x = 2011 

--> Amin = 3 <=> x = 2011 

** Kết hợp các trường hợp trên lại ta có : 

Amin = 2 <=> x = 2010 

oái,sai rồi

`Answer:`

\(\frac{2-x}{2009}-1=\frac{1-x}{2010}-\frac{x}{2011}\)

\(\Leftrightarrow\frac{2-x}{2009}+1=\frac{1-x}{2010}+1-\frac{x}{2011}+1\)

\(\Leftrightarrow\frac{2011-2}{2009}=\frac{2011-x}{2010}-\frac{x-2011}{2011}\)

\(\Leftrightarrow\left(2011-x\right)\left(\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

Mà \(\frac{1}{2010}+\frac{1}{2011}\ne\frac{1}{2009}\)

\(\Rightarrow2011-x=0\Leftrightarrow x=2011\)