\(\frac{3}{5}-\frac{-7}{10}+\frac{13}{20}=\frac{12}{20}-\frac{-14}{20}+\frac{13}{20}=\frac{12-\left(-14\right)+13}{20}=\frac{12+14+13}{20}=\frac{39}{20}< \frac{40}{20}=2\)

Vậy \(\frac{3}{5}-\frac{-7}{10}+\frac{13}{20}< 2\)

kết quả : A > B

Ta có: 
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80 

1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80) 

Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60 
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60 

và 1/61> 1/62> ... >1/79> 1/80 
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80 

Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12 

=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12

\(A>B\)

Cảm ơn các bạn nha!

Câu hỏi của Lê Tiến Cường - Toán lớp 6 - Học toán với OnlineMath

\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{10101}{10100}=\frac{2+1}{2}+\frac{6+1}{6}+\frac{12+1}{12}+...+\frac{10100+1}{10100}\)

\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{10100}\right)\)

\(A=\left(1+\frac{1}{1\times2}\right)+\left(1+\frac{1}{2\times3}\right)+\left(1+\frac{1}{3\times4}\right)+...+\left(1+\frac{1}{100\times101}\right)\)

\(A=\left(1+1+1+....+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{100\times101}\right)\)

\(A=100+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\right)\)

\(A=100+1-\frac{1}{101}=101-\frac{1}{101}< 101=B\)

\(\Rightarrow A< B\)

So easy

Ta có:

A = \(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2016}\)

\(=2.\left(\frac{1}{60.63}+\frac{1}{63.66}+...+\frac{1}{117.120}\right)+\frac{2}{2016}\)

\(=2.\frac{1}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2016}\)

\(=\frac{2}{3}.\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2016}\)

\(=\frac{2}{3}.\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2016}\)

\(=\frac{2}{3}.\frac{1}{120}+\frac{2}{2016}\)

\(=\frac{1}{180}+\frac{2}{2016}\)

B = \(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2016}\)

\(=\frac{5}{4}.\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2016}\)

\(=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2016}\)

\(=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2016}\)

\(=\frac{5}{4}.\frac{1}{80}+\frac{5}{2016}\)

\(=\frac{1}{64}+\frac{5}{2016}\)

Vì \(\frac{1}{64}>\frac{1}{180}\) và \(\frac{5}{2016}>\frac{2}{2016}\) nên B > A

Vậy B > A

ta có \(\frac{3}{7}=\frac{3\times3}{7\times3}=\frac{9}{21}\)(quy đồng tử)

So sánh \(\frac{9}{21}\)và \(\frac{9}{17}\)ta có:

\(21>17\Rightarrow\frac{9}{17}>\frac{9}{21}\Rightarrow\frac{9}{17}>\frac{3}{7}\)

VẬY: \(\frac{9}{17}>\frac{3}{7}\)

> nhé 

D = \(\frac{2^{2004}+1}{2^{2003}+1}\)=\(\frac{2^{2003}+2}{2^{2004}+2}\)

C = \(\frac{2^{2005}+3}{2^{2006}+3}\)= \(\frac{2^{2005}+2}{2^{2006}+2}\)

 Vậy C>D

mình chuyển 1 hạng tử của 3 từ bên d sang c nên ta được pt như trên

Ta có: 

\(10A=\frac{10^{2015}+20200}{10^{2015}+2020}=1+\frac{18180}{10^{2015}+2020}\)

\(10B=\frac{10^{2016}+20200}{10^{2016}+2020}=1+\frac{18180}{10^{2016}+2020}\)

Vì \(10^{2016}+2020>2^{2015}+2020\)

=> \(\frac{18180}{10^{2016}+2020}< \frac{18180}{10^{2015}+2020}\)

=> \(1+\frac{18180}{10^{2016}+2020}< 1+\frac{18180}{10^{2015}+2020}\)

=> 10B < 10A

=> B<A

\(A=\frac{10^{2014}+2020}{10^{2015}+2020}\)\(< \) \(B=\frac{10^{2015}+2020}{10^{2016}+2020}\)

chúc bạn học tốt

study well

Bài của cô bị lỗi ở dòng dưới nó không hiện thị
Em thêm cho cô nhé Bảo Thi

10B < 10A 

=> B < A