a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)

\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)

\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)

\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)

\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)

b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)

a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
    = \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
    = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)                                                          
    = \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
    = \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
    = \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
    = \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ... 

P/s : Đề sai mik sửa lại rồi : Tham khảo nhé : 

\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}-2.\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}-1+\frac{1}{2}+....+\frac{1}{100}\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

bn học nhanh thế 

mk chưa học đến đấy 

nâng cao à 

đồng gì đấy bn ơi,viết rõ dùm

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{200^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=1-\frac{1}{200}\)

\(=\frac{199}{200}\)

vậy \(\frac{99}{200}< \frac{199}{200}< 1\left(đpcm\right)\)

rồi sao

ta thấy : \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};\frac{1}{4^2}>\frac{1}{4.5};...;\frac{1}{199^2}>\frac{1}{199.200}\)

suy ra: \(M>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{199.200}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{2}-\frac{1}{200}\)

=\(\frac{100}{200}-\frac{1}{200}=\frac{99}{200}\)

=> \(M>\frac{99}{200}\)

ta cũng thấy: \(\frac{1}{2^2}

Ta xét 2 phân thức \(\frac{a^2}{a^2-100a+5000}\)và \(\frac{\left(100-a\right)^2}{\left(100-a\right)^2-100\left(100-a\right)+5000}\)(với \(a\in N\)và \(1\le a\le99\)).

Xét hiệu 2 mẫu: \(a^2-100a+5000-\left(100-a\right)^2+100\left(100-a\right)-5000\)

\(=a^2-100a-100^2+200a-a^2+100^2-100a=0.\)

Do đó 2 mẫu bằng nhau và \(\frac{a^2}{a^2-100a+5000}+\frac{\left(100-a\right)^2}{\left(100-a\right)^2-100\left(100-a\right)+5000}\)

\(=\frac{a^2+\left(100-a\right)^2}{a^2-100a+5000}=\frac{2a^2-200a+100^2}{a^2-100a+5000}=2\)

Thay a = 1, 2, 3, ..., 49 ta có:

\(\left(\frac{1^2}{1^2-100+5000}+\frac{99^2}{99^2-9900+5000}\right)+\left(\frac{2^2}{2^2-200+5000}+\frac{98^2}{98^2-9800+5000}\right)+...+\left(\frac{49^2}{49^2-4900+5000}+\frac{51^2}{51^2-5100+5000}\right)+\frac{50^2}{50^2-5000+5000}\)

\(=2.49+1=99\)

lấy cái tên NARUTO ở đâu mà hay ghê (ở trong BB phải ko)

là sao

Bạn tham khảo lời giải tại đây:

Câu hỏi của Nguyễn Kim Chi - Toán lớp 7 | Học trực tuyến

Và lưu ý lần sau gõ đề bằng công thức toán nhé.