Đặt biểu thức trên là A = 1/1.3 + 1/3.5 + 1/5.7 +...+ 1/99.101

2A= 2( 1/1.3 + 1/3.5 + 1/5.7 +...+ 1/99.101 ) 

2A= 2/1.3 + 2/3.5 + 2/ 5.7 +... 2/99.101

2A= 1-1/3 + 1/3 - 1/5 + 1/5-1/7 +...+ 1/99-1/101

2A= 1-1/101

2A= 100/101

A= 100/101 . 1/2

A= 50/101

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow1-\frac{1}{101}=\frac{100}{101}\)

\(250\times12-\left(242+302\times2,5\right)\)

\(=3000-997\)

\(=2003\)

Biểu thức 1 :

1  4/5 x 5/6 = 9/5 x 5/6 = 45 /30 = 3/2

Biểu thức 2 : 

15/16 : 11 /4 = 15/16 x 4/11 = 15/44

Tk cho mk nha ! Kết bạn với mk nhé !

\(\frac{72.62.14}{36.88.21}=\frac{9.8.2.31.7.2}{9.2.2.8.11.7.3}=\frac{31}{11.3}=\frac{31}{33}\)

\(\frac{x}{y}=\frac{-1}{2}\Leftrightarrow2x=-y\Leftrightarrow y=-2x\)

Thay y=-2x vào biểu thức cần tính sẽ ra

\(\frac{2016,2017-198}{2017.2015+1819}=\frac{\left(2015+1\right).2017-198}{2017.2015+1819}\)

                                        \(=\frac{2015.2017+2017-198}{2017.2015+1819}\)

                                        \(=\frac{2015.2017+1819}{2017.2015+1819}\)

                                        \(=1\)

Nhớ k cho mình nha

\(\frac{2016.2017-198}{2017.2015+1819}=\frac{\left(2015+1\right)2017-198}{2017.2015+1819}=\frac{2015.2017+2017-198}{2017.2015+1819}=\frac{2017.2015+1819}{2017.2015+1819}=1\)

B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)

B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)

B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x 0

B = 0

nhân 4 vào tử

\(\frac{1}{6.10}\)+ \(\frac{1}{10.14}\)+ ... + \(\frac{1}{402.406}\)

= \(\frac{1}{4}\). \(\left(\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\right)\)

= \(\frac{1}{4}\). ( \(\frac{10-6}{6.10}\)+ \(\frac{14-10}{10.14}\)+ ... + \(\frac{406-402}{402.406}\))

= \(\frac{1}{4}\). ( \(\frac{10}{6.10}\)- \(\frac{6}{6.10}\)+ ... + \(\frac{406}{402.406}\)- \(\frac{402}{402.406}\))

= \(\frac{1}{4}\). ( \(\frac{1}{6}\)- \(\frac{1}{406}\))

= \(\frac{1}{4}\). \(\frac{100}{609}\)

= \(\frac{25}{609}\)