\(a,A=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)

\(A=\frac{1}{2}\left[\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+...+\frac{2}{73\cdot75}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{75}\right]=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(b,B=\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+...+\frac{1}{197\cdot200}\)

\(3B=\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\)

\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\)

\(3B=\frac{1}{8}-\frac{1}{200}\)

\(3B=\frac{3}{25}\)

\(B=\frac{3}{25}:3=\frac{1}{25}\)

#)Giải :

a, \(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(A=\frac{1}{25}-\frac{1}{75}\)

\(A=\frac{2}{75}\)

b, \(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)

\(B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)

\(B=\frac{1}{8}-\frac{1}{200}\)

\(B=\frac{3}{25}\)

            #~Will~be~Pens~#

\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(2A=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)

\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(2A=\frac{1}{25}-\frac{1}{75}\)

\(2A=\frac{2}{75}\)

\(A=\frac{2}{75}:2=\frac{1}{75}\)

\(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)

\(3B=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)

\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)

\(3B=\frac{1}{8}-\frac{1}{200}\)

\(3B=\frac{3}{25}\)

\(B=\frac{3}{25}:3=\frac{1}{25}\)

Bài làm

c ) Ta có :

 \(\frac{2017}{2018}< 1\)

\(\frac{12}{11}>1\)

\(\Rightarrow\frac{2017}{2018}< \frac{12}{11}\)

trả lời

a, quy đồng rồi so sánh 

b,quy đồng rồi so sánh 

c,phân số nào có tử nhỏ hơn mẫu khi so sành với phân số có tử lớn hơn mẫu đều bé hơn

d,quy đồng rồi so sánh

chắc vậy chúc bn học tốt

Bài làm

b ) Ta có :

\(\frac{13}{27}=\frac{26}{54}\)

Mà : \(\frac{26}{54}< \frac{27}{54}\)

Lại có : \(\frac{27}{54}< \frac{27}{41}\)

\(\Rightarrow\frac{13}{27}< \frac{27}{41}\)

\(\left(1+\frac{1}{4}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right).\left(1+\frac{1}{24}\right)...\left(1+\frac{1}{9999}\right)\)

\(=\frac{5}{4}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{10000}{9999}=\frac{5.9.16.25...10000}{4.8.15.24...9999}=\frac{5.3^2.4^2.5^2...100^2}{4.2.4.3.5.4.6...99.101}\)

\(=\frac{5.3.4.5...100.3.4.5...100}{4.2.3.4...99.4.5.6...101}=\frac{5.100.3}{4.2.101}=\frac{5.25.3}{2.101}=\frac{375}{202}.\)

1/2 . P = 1/2.6 + 1/6.10 + 1/10.14 + ... + 1/198.202

4.1/2. P= 4/2.6 + 4/6.10 + 4/10.14 + ... + 4/198.202

2P=1/2-1/6+1/6-1/10+1/10-1/14+...+1/198-1/202

2P=1/2-1/202=50/101

P=50/101:2=50/101.1/2=25/101

\(\frac{\frac{4}{17}-\frac{4}{45}+\frac{4}{156}}{\frac{3}{17}-\frac{3}{45}+\frac{3}{156}}=\frac{4.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}{3.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}=\frac{4}{3}\)

thanks Le Tai Bao Chau nha

\(\frac{\frac{4}{17}-\frac{4}{45}+\frac{1}{39}}{\frac{3}{17}-\frac{1}{15}+\frac{1}{52}}\)

\(=\frac{\frac{4}{17}-\frac{4}{45}+\frac{4}{156}}{\frac{3}{17}-\frac{3}{45}+\frac{3}{156}}\)

\(=\frac{\frac{1}{17}.4-\frac{1}{45}.4+\frac{1}{156}.4}{\frac{1}{17}.3-\frac{1}{45}.3+\frac{1}{156}.3}\)

\(=\frac{4\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}{3\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}\)

\(=\frac{4}{3}\)

\(\frac{9}{11}\)nha kb đi

lấy (1/3 + 1/15 +1/10 + 1/21 ) + (1/36 + 1/28 + 1/6) + (1/45 + 1/55)

      =  (4/50 + 3/70) + 2/100

      = 7/120 + 2/100

      = 9/220

= 9/11

\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\) 

\(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(\frac{1}{2}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}A=1-\frac{1}{10}\)

\(\frac{1}{2}A=\frac{9}{10}\)

\(A=\frac{9}{10}:\frac{1}{2}\)

\(A=\frac{18}{10}=\frac{9}{5}\)

sao thế tự dưng lấy 1/12 ở đâu thế 1/10 cơ mà

đề bài thiếu = 2/9

1/21+1/28+1/36+...+2/x(x+1)=2/9

=>1/3.7+1/4.7+...+2/x.(x+1)=2/9

=>2/6.7+2/7.8+...+2/x.(x+1)=2/9

=>2.(1/6-1/x+1)=2/9

=>1/6-1/x=1=1/9

=>1/x+1=1/18

=>x+1=18

=>x=17

CHÚC BẠN HỌC GIỎI

K MÌNH NHÉ

\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)

\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)

\(\left(2x-1\right)=-\frac{4}{63}\)

2x= -4/63 + 1

2x = 59/63

x = 59/63 : 2

x = 59/126

1/3:(2.x-1)=-5-1/4

1/3:(2.x-1)=-21/4

2.x-1=1/3:-21/4

2.x-1=-4/63

2.x=-4/63+1

2.x=\(3\frac{59}{63}\)

x=\(3\frac{59}{63}\):2

x=\(1\frac{61}{63}\)

\(\frac{1}{4}+\frac{1}{3}\div\left(2x-1\right)=-5\)

\(\frac{1}{3}\div\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-20}{4}+\frac{-1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-21}{4}\)

\(2x-1=\frac{1}{3}\div\frac{-21}{4}\)

\(2x-1=\frac{1}{3}.\frac{-4}{21}\)

\(2x-1=\frac{-4}{63}\)

\(2x=\frac{-4}{63}+1\)

\(2x=\frac{-4}{63}+\frac{63}{63}\)

\(2x=\frac{59}{63}\)

\(x=\frac{59}{63}\div2\)

\(x=\frac{59}{63}.\frac{1}{2}\)

\(x=\frac{59}{126}\)