A = 1 - 1/2 + 1/3 - 1/4 + ..... + 1/19 - 1/20

   = 1 + 1/2 + 1/3 + 1/4 +............+ 1/19 + 1/20 - 2(1/2 + 1/4 + 1/6 +...........+ 1/20)

   = 1 + 1/2 + 1/3 + 1/4 +............+ 1/19 + 1/20 - ( 1 + 1/2 + 1/3 + .............. + 1/10)

   = 1/11 + 1/12 + ...... + 1/20

=> A / B = 1

Ta có:

A = \(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.5+15.20}\)(sửa 6,8 = 6 . 8 hay  6 x  8)

Ở phép tính này. Chúng ta thực hiện lượt các số giống nhau đi

Ta lại được:

\(\frac{1.2+2+3+4+5}{3+9.12+12+15.20}\)=  \(\frac{16}{423}\)

Rồi thực hiện so sánh. Tự làm nhá!

\(A=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

\(A=\frac{1.2.\left(1+2^2+3^2+4^2+5^2\right)}{3.4.\left(1+2^2+3^2+4^2+5^2\right)}\)

\(A=\frac{1.2}{3.4}\)

\(A=\frac{1}{6}\)

Ta thấy : \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)

Vậy B > A

Theo đề bài, ta có:

\(A=\frac{1\times2+2\times4+3\times6+4\times8+5\times10}{3\times4+6\times8+9\times12+12\times16+15\times20}\)

\(A=\frac{1\times2\times\left(1+2^2+3^2+4^2+5^2\right)}{3\times4\times\left(1+2^2+3^2+4^2+5^2\right)}\)

\(A=\frac{1\times2}{3\times4}\)

\(A=\frac{1}{6}\)

Ta thấy rằng: \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)

Vậy \(B>A\)

ai giúp mk vs ạ

dở lại sách lớp 6

bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó

CÂU 1 = -59/111

 CÂU 2 = 11/63

cảm ơn kết quả thì mik b òi nhưng mik cần cách làm

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)

\(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75}\), \(\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên :

\(A>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(A< \frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{7}{12}< A< \frac{5}{6}\)

+A=1/(1.2)+1/(3.4)+...+1/(99.100) 
=1/1-1/2+1/3-1/4+....+1/99-1/100 
=1/2+1/3-1/4+1/5-1/6+1/7+...-1/98+1/99... 
=(1/2+1/3)+(1/5-1/4)+(1/7-1/6)+..+(1/9... 
=5/6-(1/4.5+1/6.7+..1/98.99+1/100)<5/6 
do -(1/4.5+1/6.7+..1/98.99+1/100)<0 
+A=1/(1.2)+1/(3.4)+...+1/(99.100) 
=1/2+1/12+1/(5.6)+...+1/(99.100) 
=7/12+[1/(5.6)+...1/(99.100)] 
>7/12 do [1/(5.6)+...1/(99.100)]>0

1-1/100=99/100

7/12<2/3<5/6

A=1/1x2 +1/2x3 +... +1/18x19 + 1/19x20

Nhận xét 1/1x2 = 1/1 -1/2 ; 1/2x3=1/2-1/3; ... ;1/18x19=1/18-1/19 ; 1/19x20=1/19-1/20

ta có A=1/1 - 1/2 + +1/2 -1/3+1/3-   +1/18-1/19+1/19-1/20

         A=1/1 - 1/20 

         A=20/20 - 1/20

         A=(20-1)/20

         A=19/20

                   Vậy A=19/20

A =\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+ ...+\(\frac{1}{18.19}\)+\(\frac{1}{19.20}\)

A = 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+....+\(\frac{1}{18}\)- \(\frac{1}{19}\)+ \(\frac{1}{19}\)- \(\frac{1}{20}\)

A = 1 - \(\frac{1}{20}\)( Vì đã triệt tiêu )

A = \(\frac{19}{20}\)

a=(1/1-1/2) + (1/2 +1/3) +(1/3+1/4)+........+(1/2016+1/2017)

a=1/1-1/2+1/2-1/3+1/3-1/4+ ........+2016-2017

a=1/1+1/2017 =2017-1/2017

a=2016/2017

chac chan 100% do ban .tk mk nha