\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=2018\)

Ta có : 

\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)

\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)

\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)

\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)

\(A=2018\)

Vậy \(A=2018\)

Chúc bạn học tốt ~ 

mình giỏi vl vậy >:

Còn cần không:v

Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)

=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)

=>\(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2017}}\)

\(A=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^3}-\frac{1}{2^3}\right)+...+\left(\frac{1}{2^{2016}}-\frac{1}{2^{2016}}\right)-\frac{1}{2^{2017}}\)

\(A=1-\frac{1}{2^{2017}}\)

Vậy: \(A=1-\frac{1}{2^{2017}}\)

\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)

\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)

\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)

\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)

3/\(7a+b=0\Rightarrow b=-7a\)

\(f\left(x\right)=ax^2-7ax+c\).Ta có: \(f\left(10\right)=100a-70a+c=30a+c\)

\(f\left(-3\right)=30a+c\).Nhân theo vế ta có đpcm:

\(f\left(10\right).f\left(-3\right)=\left(30a+c\right)^2\ge0\) (đúng)

Câu 2 tự quy đồng rồi so sánh:v

C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!