a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}=\frac{10^{2014}+10}{10^{2015}+10}=\frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2013}+1}{10^{2014}+1}=A\)

\(\Rightarrow\)\(B< A\) hay \(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~ 

áp dụng tính chất

nếu a/b>1thì a/b<(a+n)/(b+n)

=)))))))))))))))))

Vì \(B=\frac{2014^{11}+2}{2014^{12}+2}<1\)

\(\Rightarrow B=\frac{2014^{11}+2}{2014^{12}+2}<\frac{2014^{11}+2+4026}{2014^{12}+2+4026}=\frac{2014^{11}+4028}{2014^{12}+4028}=\frac{2014.\left(2014^{10}+2\right)}{2014\left(2014^{11}+2\right)}=\frac{2014^{10}+2}{2014^{11}+2}=A\)

Vậy B<A hay A<B

ta chứng minh bài toán phụ:

nếu ta có b<d \(\frac{a}{b}\)>\(\frac{c}{d}\) thì ad>bc

dễ thây \(\frac{ad}{bd}>\frac{cb}{bd}\)

 => ad>bd

áp dụng:

dat 2014=a ta co

\(A=\frac{a^{10}+2}{a^{11+2}}\)

 \(B=\frac{a^{11}+2}{a^{12}+2}\)

 ta có 

\(A=\frac{a^{10}+2.a^{12}+2}{a^{11}+2.a^{12}+2}\)

 \(B=\frac{a^{11}+2.a^{11}+2}{a^{12}+2.a^{11}+2}\)=\(\frac{a^{10}+2a^{12}+2}{a^{12}+2a^{11}+2}\)

 => A=B

mk hok chắc đâu nha

HELP ME!!!!!!!!!!!!!!

4/ Vì 4x/6y=2x+8/3y+11

\(\Rightarrow\)  4x(3y+11)=6y(2x+8)

\(\Rightarrow\)  12xy+44xy= 12xy+48xy

\(\Rightarrow\)   44xy= 48xy

\(\Rightarrow\frac{x}{y}=\frac{48}{44}=\frac{12}{11}\)

dit me may

Người lái xe trước khi đi thấy chỉ còn 3/5 thùng xăng, sợ không đủ nên người đó mua thêm 14 lít xăng nữa. Khi về tới nhà anh thấy chỉ còn 1/3 thùng xăng và tính ra xe tiêu thụ hết 30 lít xăng trong chuyến đi đó. Hỏi thùng xăng chứa bao nhiêu lít xăng?

Bài 1 : 

Ta có :

\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)

Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)

Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)

Vậy \(A>B\)

Bài 2 :

Ta có :

\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)

\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)

\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)

\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)

Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên  \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)

Nên : \(M>4\)

Vậy \(M>4\)

Bài 3 : 

Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)

Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)

\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)

\(\Rightarrow A< \frac{3}{4}\)

Vậy \(A< \frac{3}{4}\)

Bài 4 :

\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)

\(\Rightarrow A=\frac{1008}{2017}\)

Vậy \(A=\frac{1008}{2017}\)

\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)

\(1-\frac{1}{x+2}=\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)

\(\Rightarrow x+2=2017\)

\(\Rightarrow x=2017-2=2015\)

Vậy \(x=2015\)