\(A=\left(-\frac{5}{11}\right).\frac{7}{15}+\frac{11}{-5}.\frac{30}{33}\)

\(A=-\frac{7}{33}+-2\)

\(A=-\frac{73}{33}\)

[ A] = -2

làm đc hết rùi phần b thui

\(D=\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\left(\frac{1}{4}+\frac{3}{8}\right)\right]\)

\(D=\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\frac{5}{8}\right]\)

\(D=\frac{1}{3}-\left(-\frac{15}{8}\right)\)

\(D=\frac{53}{24}\)

\(\left[D\right]=2\)

các giá trị tuyệt đối trên có tổng lớn hơn hoặc bằng 0(>=0)

=>100x>=0

=>x>=0 =>x+1/(1.2) >0 ;x+1/(2.3)>0;x+1/(3.4);.....;x+1/(99.100)>0

=> ta có thể phá dấu giá trị tuyệt đối 

=>100x=x+x+...+x(có 99. x)+(1/(1.2)+1/(2.3)+..+1/(99.100))

=>100x=99x+99/100

=>x=99/100

\(P=1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

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

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

     \(=2-\frac{1}{n+1}=\frac{2\left(n+1\right)}{n+1}-\frac{1}{n+1}=\frac{2n+2-1}{n+1}=\frac{2n+1}{n+1}\)

\(P=1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{n\left(n+1\right)}=1+1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{\left(n+1\right)}\)

\(\Rightarrow P=2-\frac{1}{\left(n+1\right)}=\frac{2n+1}{n+1}\)

Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge0\)

Từ điều kiện trên ta có :

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)

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

\(50x=1-\frac{1}{100}\)

\(50x=\frac{99}{100}\)

\(x=\frac{99}{5000}\)

Do \(\left|a\right|\ge0\forall a\) nên:

\(A=\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\forall x\)

\(\Leftrightarrow100x\ge0\) hay \(x\ge0\)

Do vậy ta có: \(A=\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\) ( 50 chữ số x)

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

\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)

\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)

tưởng là có 99x lận mà

\(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\) 

\(\Rightarrow a=\frac{2020}{2019}\)

=.> 1-1/2+1/2-1/3+.......+1/2019-1/2020=1/x

=>1-1/2020=1/x

=>2019/2020=1/x

=>2019x=2020

=>x=2020/2019

    k nha

 giúp mk lên 300sp

\(\frac{1}{1.2}+\frac{1}{2.3}+.........\frac{1}{2019.2020}\)

\(\Rightarrow\frac{1}{2}+\frac{1}{2}+\frac{1}{3}++........\frac{1}{2020}\)

\(\Rightarrow\frac{2019}{2020}\)

vậy \(\frac{1}{x}=x=\frac{2020}{2019}\)

h.ọ.c t.ố.t

đề chưa đầy đủ

à đề thiếu tổng các giá trị tuyệt đối ở trên =100x

What wrong with you ?