(-1)+3+(-5)+7+...+x=600

<=>[(-1)+3]+[(-5)+7]+....+[(-x)-2]+x]=600

Ta có 2+ 2 + .... + 2 = 600

=> 1 + 1 + .... + 1 = 300 

Số dấu ngoặc [] là :  \(\frac{x-3}{4}\)+ 1 

=>  \(\frac{x-3}{4}\)+ 1 = 300

=>  \(\frac{x-3}{4}\)= 299

=> x - 3 = 299 . 4 = 1199

Vậy x = 1199 

# Học Tốt

Tk cho mình nhé !

Đây là dãy số nguyên tố nên không có quy luật.
11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271,...

 vì đây là dãy số nguyên tố nên k có quy luật cụ thể....., 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271,275....

246286444728642849

a, Áp dụng định lý Ta-lét ta có:
\(\dfrac{AD}{DB}=\dfrac{AE}{EC}\Rightarrow\dfrac{4}{x}=\dfrac{5}{10}\Rightarrow x=4:\dfrac{1}{2}\Rightarrow x=8\)

Áp dụng hệ quả định lý Ta-lét ta có:

\(\dfrac{AE}{AC}=\dfrac{DE}{BC}\Rightarrow\dfrac{5}{15}=\dfrac{6}{y}\Rightarrow y=6:\dfrac{1}{3}\Rightarrow y=18\)

b, Áp dụng định lý phân giác ta có:
\(\dfrac{DB}{DC}=\dfrac{AB}{AC}\Rightarrow\dfrac{5}{6}=\dfrac{10}{x}\Rightarrow x=10:\dfrac{5}{6}\Rightarrow x=12\)

???

what

where is câu hỏi??????

1. What did the people do when you were there

2. I visited Dalat with my parents.

3. What do you think of Tam?

4. The life in countryside is peaceful and more than I expected

1.What did the people do when you were there?

2.I visited Da Lat with my parents.

3.What do you think of Tam?

4.The life in the countryside is more peaceful than I expected.

Ta có \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{49.51}\)

         =\(\dfrac{2}{2}\).(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{49.51}\))

         =\(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{49.50}\))

         =\(\dfrac{1}{2}\).(1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\))

         =\(\dfrac{1}{2}\).(\(1-\dfrac{1}{51}\))

         =\(\dfrac{1}{2}\).\(\dfrac{50}{51}\)

         =\(\dfrac{25}{51}\)

Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{49\cdot51}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{49\cdot51}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{51}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{50}{51}=\dfrac{25}{51}\)