Trả lời:

\(a,\)\(A=\frac{1990.32-990}{1990.31+1000}\)

\(\Leftrightarrow A=\frac{\left(1000+990\right).32-990}{\left(1000+990\right).31+1000}\)

\(\Leftrightarrow A=\frac{1000.32+990.32-990}{1000.31+990.31+1000}\)

\(\Leftrightarrow A=\frac{1000.32+990.31}{1000.32+990.31}\)

\(\Leftrightarrow A=1\)

Vậy\(A=1\)

\(b,\)\(B=\frac{2008.2009+2000}{2009.2010-2018}\)

\(\Leftrightarrow B=\frac{2008.2009+2008-8}{2009.2010-2010-8}\)

\(\Leftrightarrow B=\frac{2008.\left(2009+1\right)-8}{(2009-1).2010-8}\)

\(\Leftrightarrow B=\frac{2008.2010-8}{2008.2010-8}\)

\(\Leftrightarrow B=1\)

Vậy\(B=1\)

Hok tốt!

Good girl

Trả lời:

\(a,\)\(A=\frac{1990.32-990}{1990.31+1000}\)

\(\Leftrightarrow A=\frac{\left(1000+990\right).32-990}{\left(1000+990\right).31+1000}\)

\(\Leftrightarrow A=\frac{1000.32+990.32-990}{1000.31+990.31+1000}\)

\(\Leftrightarrow A=\frac{1000.32+990.31}{1000.32+990.31}\)

\(\Leftrightarrow A=1\)

Vậy\(A=1\)

\(b,\)\(B=\frac{2008.2009+2000}{2009.2010-2018}\)

\(\Leftrightarrow B=\frac{2008.2009+2008-8}{2009.2010-2010-8}\)

\(\Leftrightarrow B=\frac{2008.\left(2009+1\right)-8}{(2009-1).2010-8}\)

\(\Leftrightarrow B=\frac{2008.2010-8}{2008.2010-8}\)

\(\Leftrightarrow B=1\)

Vậy\(B=1\)

Hok tốt!

Bad boy

Ta có: 2009.2010>2008.2009

\(\frac{1}{2009\cdot2010}< \frac{1}{2008\cdot2009}\)

\(\Rightarrow E>F\)

2008/2008.2009 và 2009/2009.2010

2008/2008.2009 < 2009/2009.2010

k mk na <3

2008/ 2008 × 2009> 2009/ 2009 × 2010

Mình thề 100% CHUẨN KHÔNG CẦN CHỈNH☺

a, \(\dfrac{2009}{2010}\) và \(\dfrac{2010}{2011}\)

Ta có:

\(2009.2011=4040099\)

\(2010.2010=4040100\)

Vì \(2009.2011< 2010.2010\)

nên \(\dfrac{2009}{2010}< \dfrac{2010}{2011}\)

b, \(\dfrac{2008}{2008.2009}\) và \(\dfrac{2009}{2009.2010}\)

Ta có:

\(\dfrac{2008}{2008.2009}=\dfrac{1}{2009};\dfrac{2009}{2009.2010}=\dfrac{1}{2010}\)

Vì \(\dfrac{1}{2009}>\dfrac{1}{2010}\) nên \(\dfrac{2008}{2008.2009}>\dfrac{2009}{2009.2010}\)

Chúc bạn học tốt!!!

a)\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(\dfrac{2009}{2010}< 1\)

\(\Leftrightarrow\dfrac{2009}{2010}< \dfrac{2009+1}{2010+1}\Leftrightarrow\dfrac{2009}{2010}< \dfrac{2010}{2011}\)

b)

\(\dfrac{2008}{2008.2009}=\dfrac{1}{2009}\)

\(\dfrac{2009}{2009.2010}=\dfrac{1}{2010}\)

\(\dfrac{1}{2009}>\dfrac{1}{2010}\Leftrightarrow\dfrac{2008}{2008.2009}>\dfrac{2009}{2009.2010}\)

d)

\(\dfrac{1}{3^{400}}=\dfrac{1}{\left(3^4\right)^{100}}=\dfrac{1}{81^{100}}\)

\(\dfrac{1}{4^{300}}=\dfrac{1}{\left(4^3\right)^{100}}=\dfrac{1}{64^{100}}\)

\(81^{100}>64^{100}\Leftrightarrow\dfrac{1}{81^{100}}< \dfrac{1}{64^{100}}\)

a, Lấy 1 trừ từng phân số.

\(1-\dfrac{2009}{2010}=\dfrac{1}{2010}\)

\(1-\dfrac{2010}{2011}=\dfrac{1}{2011}\)

Vì \(\dfrac{1}{2010}>\dfrac{1}{2011}\) nên \(\dfrac{2009}{2010}< \dfrac{2010}{2011}\).

b, \(\dfrac{2008}{2008.2009}=\dfrac{1}{2009}\)

\(\dfrac{2009}{2009.2010}=\dfrac{1}{2010}\)

Vì \(\dfrac{1}{2009}>\dfrac{1}{2010}\) nên \(\dfrac{2008}{2008.2009}>\dfrac{2009}{2009.2010}\)

c, Ta có:

\(\dfrac{2016}{2017}>\dfrac{2016}{2018}\Rightarrow\dfrac{2016}{2017}+\dfrac{2017}{2018}>\dfrac{2016}{2018}+\dfrac{2017}{2018}\)

\(\Rightarrow\dfrac{2016}{2017}+\dfrac{2017}{2018}>\dfrac{2016+2017}{2018}\)

\(\dfrac{2016+2017}{2017+2018}=\dfrac{2016+2017}{4035}\)

Vì \(\dfrac{2016+2017}{2018}>\dfrac{2016+2017}{4035}\) nên \(\dfrac{2016}{2017}+\dfrac{2017}{2018}>\dfrac{2016+2017}{2017+2018}\)

d, \(\left(\dfrac{1}{3}\right)^{400}=\left(\dfrac{1}{3}\right)^{4^{100}}=\left(\dfrac{1}{81}\right)^{100}\)

\(\left(\dfrac{1}{4}\right)^{300}=\left(\dfrac{1}{4}\right)^{3^{100}}=\left(\dfrac{1}{64}\right)^{100}\)

Vì \(\left(\dfrac{1}{81}\right)^{100}< \left(\dfrac{1}{64}\right)^{100}\) nên \(\left(\dfrac{1}{3}\right)^{400}< \left(\dfrac{1}{4}\right)^{300}\)

a/ Do : 2009/2010 > 2009/2011, 2009/2011 < 2010/2011 nên 2009/2010 < 2010/2011

                                   1 đúng

Ta có: 200/201+201/202>200+201/202          (1)

200+201/201+202<200+201/202                   (2)

từ (1) và (2) suy ra 200/201+201/202>200+201/201+202

biết làm câu đ ko bạn Đỗ Hà Khánh Linh

a,

\(2018^2-2017\cdot2019\\ =2018^2-\left(2018-1\right)\left(2018+1\right)\\ =2018^2-2018^2+1\\ =1\)

b, Đề khó nhìn bạn ạ, gõ Latex đi bạn! :)

a) 2018^2-2017.2019

= 2018^2 -(2018-1)(2018+1)

= 2018^2 -2018^2 -1

=-1

a.\(2018^2-2017\cdot2019\)

\(=2018^2-\left(2018-1\right)\cdot\left(2018+1\right)\)

\(=2018^2-\left(2018^2-1\right)\)

\(=2018^2-2018^2+1\)

=1

b.\(2018^3+\frac{1}{2018-2017}\)

\(=2018^3+1\)

nếu thích thì hằng đẳng thức cho vui

dễ mà bạn làm từ câu a nếu ra thì các câu khác cũng dễ thôi

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A=1-\frac{1}{2010}\)

\(A=\frac{2009}{2010}\)