2009.2009=2009(2008+1)=2009.2008+2009

2008+2010=2008(2009+1)=2008.2009+2008

=> 2009.2009>2008.2010

k hộ mình cái

B = 2008.2010 = (2009 - 1).(2009 + 1) = 2009.(2009 + 1) - 1.(2009 + 1) = 2009.2009 + 2009 - 2009 - 1 = 2009.2009 - 1 < A

Vậy A > B

A=\(\frac{2008.\left(2009+1\right)+447}{\left(2008+1\right).2009+476}\)=\(\frac{2008.2008+2008+447}{2008.2009+2009+446}\)=1

Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)

\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)

\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)

\(\frac{2009.2009+2008}{2009.2009+2009}=\frac{2009.2009+2009}{2009.2009+2009}-\frac{1}{2009.2009+2009}=1-\frac{1}{2009.2009+2009}\)

\(\frac{2009.2009+2009}{2009.2009+2010}=\frac{2009.2009+2010}{2009.2009+2010}-\frac{1}{2009.2009+2010}=1-\frac{1}{2009.2009+2010}\)

\(\text{Vì }2009.2009+2009\frac{1}{2009.2009+2010}\)

\(\text{Hay }1-\frac{1}{2009.2009+2009}

\(\frac{2009.2009+2009}{2009.2009+2010}=\frac{2009.2009+2008+1}{2009.2009+2009+1}\)

Đặt 2009.2009+2008 là a; 2009.2009+2009 là b. Ta so sánh \(\frac{a}{b}\)và \(\frac{a+1}{b+1}\)

Qui đồng mẫu số 2 phân số trên

\(\frac{a}{b}=\frac{a\left(b+1\right)}{b\left(b+1\right)}=\frac{a.b+a}{b.\left(b+1\right)}\)

\(\frac{a+1}{b+1}=\frac{\left(a+1\right).b}{b\left(b+1\right)}=\frac{a.b+b}{b\left(b+1\right)}\)

Vì 2008 < 2009

=> 2009.2009+2008 < 2009.2009+2009

=> a < b

=> a.b+a < a.b+b

=> \(\frac{a.b+a}{b.\left(b+1\right)}

Ta có :

\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+..................+\dfrac{4}{2008.2010}\)

\(\Rightarrow F=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.............+\dfrac{2}{2008.2010}\right)\)

\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(\Rightarrow F=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)

\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+......+\dfrac{4}{2008.2010}\)

\(F=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{2008.2010}\right)\)

\(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)\(F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)\(F=2.\dfrac{502}{1005}\)

\(F=\dfrac{1004}{1005}\)

Trước hết ta chứng minh (a-1)(a+1) + 1 = a^2 (*)

Thật vậy VT = (a-1)(a+1)+1=(a-1)a + a-1 +1 = a^2-a+a=a^2 =VP 

Áp dụng (*) ta có:

\(A=\dfrac{1\cdot3+2}{2^2}+\dfrac{2\cdot4+2}{3^2}+...+\dfrac{2009\cdot2011+2}{2010^2}\\ =\dfrac{2^2+1}{2^2}+\dfrac{3^2+1}{3^2}+...+\dfrac{2010^2+1}{2010^2}=2009+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2010^2}\\ < 2009+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2009\cdot2010}\\ =2009+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2009}-\dfrac{1}{2010}=2010-\dfrac{1}{2010}< 2020< 2011\)

a)\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2008\cdot2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

b)\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

a) gọi biểu thức đó là A

Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)

Dựa vào công thức trên, ta có

\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)

\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2009}\right)\)

\(A=2.\left(\dfrac{2007}{4018}\right)=\dfrac{2007}{2009}\)

b) dễ quá bạn tự làm. (không phải mink không biết làm đâu nha)