Đáp án:\(\frac{47}{30}\)

Ta có \(\left|x+1\right|\ge0;\left|x+\frac{1}{3}\right|\ge0;...;\)\(\left|x+\frac{1}{190}\right|\ge0\)    \(\forall x\)

=> \(\left|x+1\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{190}\right|\ge0\)   \(\forall x\)

=> \(20x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\)  =>  \(x+1>0,x+\frac{1}{3}>0,x+\frac{1}{6}>0,...,x+\frac{1}{190}>0\)

                        => \(\left|x+1\right|=x+1,\left|x+\frac{1}{3}\right|=x+\frac{1}{3},\left|x+\frac{1}{6}\right|=x+\frac{1}{6},...,\left|x+\frac{1}{190}\right|=x+\frac{1}{190}\)

=> \(x+1+x+\frac{1}{3}+x+\frac{1}{6}+...+x+\frac{1}{190}=20x\)

=> \(19x+\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)=20x\)

=> \(x=\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)\)

Gọi \(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{190}\)

=> \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)

=> \(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

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

=> \(\frac{1}{2}A=1-\frac{1}{20}\)

=> \(A=\frac{19}{10}\)

Thay vào ta có 

=> \(x=-\frac{19}{10}\)

mk nhầm nha bạn \(x=\frac{19}{10}\)

Ta co:

\(\frac{1}{2}A=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}+\frac{1}{90}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}+\frac{1}{9.10}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{10}=\frac{13}{20}\Rightarrow A=\frac{13}{10}.\)

\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{36}+\frac{1}{45}\)

\(A=\frac{2}{4}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{72}+\frac{2}{90}\)

\(A=\frac{2}{2.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{8.9}+\frac{2}{9.10}\)

\(A=2\left(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(A=2.\frac{2}{5}\)

\(A=\frac{4}{5}\)

~ Học tốt ~ K cho mk nhé! Thank you.

A = -1 - 1/3 - 1/6 - 1/10 - 1/15 - ... - 1/1225

A = -(1 + 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/1225)

A = -(2/2 + 2/6 + 2/12 + 2/20 + 2/30 + ... + 2/2450)

A = -2.(1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/49.50)

A = -2.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/49 - 1/50)

A = -2.(1 - 1/50)

A = -2.49/50

A = -49/25

a) =\(\frac{1}{2}.\frac{2}{3}.....\frac{2017}{2018}=\frac{1.2.....2017}{2.3.4.....2018}=\frac{1}{2018}\)

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

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

\(=\frac{1.2...2017}{2.3...2018}\)

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

b) \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{190}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{189}{190}\)

\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{378}{380}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{7.4}{5.6}...\frac{18.21}{19.20}\)

\(=\frac{\left(1.2.3...18\right).\left(4.5.6...21\right)}{\left(2.3.4...19\right).\left(3.4.5...20\right)}\)

\(=\frac{1.21}{19.3}\)

\(=\frac{21}{57}\)

c) \(\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)\left(1+\frac{7}{48}\right)...\left(1+\frac{7}{2009}\right)\)

\(=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.\frac{56}{48}...\frac{2016}{2009}\)

mk ko bít làm câu c ! xin lỗi bn nha! bn tự nghĩ cách làm câu c giúp mk nhé!

\(P=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{189}{190}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}........\frac{378}{380}\)

\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}........\frac{18.21}{19.20}=\frac{1.2.3......18}{2.3.4....19}.\frac{4.5.6....21}{3.4.5....20}\)

\(P=\frac{1}{19}.\frac{21}{3}=\frac{21}{57}\)