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

=\(\frac{1}{2018}\)

bạn trừ ra là đc

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)

\(=\frac{1\cdot2\cdot3\cdot....\cdot2016\cdot2017}{2\cdot3\cdot4\cdot....\cdot2017\cdot2018}\)

\(=\frac{1}{2018}\)

dấu lớn hơn vì 23/12 lớn hơn 2008/2009

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

\(=\frac{1.2.3....2017}{2.3.4...2018}\)

\(=\frac{1}{2018}\)

\(=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2018}\right)\)  

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

\(=\frac{1}{2018}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)

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

\(=\frac{1.2.3.4...2016.2017}{2.3.4...2017.2018}=\frac{1}{2018}\)

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}=\frac{-5}{x\left(x-5\right)}\)

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}\)

\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)

\(=\frac{x-5-x}{x\left(x-5\right)}\)

\(=-\frac{5}{x\left(x-5\right)}\)

\(=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)

1/ (x+1)(x+2) +1/ (x+2)(x+3) +1/ (x+3)(x+4) +1/ (x+4)(x+5)

=1/x+1 -1/x+2 +1/x+2 -1/x+3 +1/x+3 -1/x+4 +1/x+4 -1/x+5

=1/x+1 -1/x+5

=4/(x+1)(x+5)

\(=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)

\(=\frac{1}{x+1}-\frac{1}{x+5}=\frac{x+5-x-1}{\left(x+1\right)\left(x+5\right)}=\frac{4}{\left(x+1\right)\left(x+5\right)}\)

Ta có:

\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)

\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)

\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)

\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)

\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)

b/ Thế vô rồi tính nhé

Đoạn gần cuối thay y-x= 1 luôn 

\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)

\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)

\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\)  giờ mới thay không biết đã tối giản chưa

MTC: (x+y)(x+1)(1-y)

\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)

\(=x-y+xy\)

Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định