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2) = -1/2 . -2/3 .-3/4 ..... . -98/99 = 1/99 (Tích này có 98 thừa số âm, 98 là số chẵn nên tích mang dấu dương)
câu 1
\(\Leftrightarrow A=\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+...+\frac{4}{107}-\frac{4}{111}\)
\(\Rightarrow A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(\Rightarrow A=4.\left(\frac{1}{3}-\frac{1}{111}\right)\)
\(\Rightarrow A=4.\frac{12}{37}\)
\(\Rightarrow A=\frac{48}{37}\)
phần B làm tương tự
câu 2:
a)\(\Leftrightarrow x+\frac{7}{12}=\frac{15}{18}\)
\(\Rightarrow x=\frac{15}{18}-\frac{7}{12}\)
\(\Rightarrow x=\frac{1}{4}\)
b,c tương tự như câu 1 phần a
Câu 1:
Ta có: A=1/3-1/7+1/7-1/11+....+1/107-1/111
=> A=1/3+(-1/7+1/7)+(-1/11+1/11)+....+(-1/107+1/107)+(-1)/111
=>A=1/3+(-1)/111
=>A=12/37
Ta có B= 6(1/15.18+1/18.21+...+1/87.90)
=> 3B= 6(3/15.18+3/18.21+...+3/87.90)
=> 3B= 6(1/15-1/18+1/18-1/21+....+1/87-1/90)
(Tương tự như câu A) 3B=6[1/15+(-1)/90]
=> 3B= 6.1/18=1/3
=> B= 1/3:3 = 1/9
a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)
\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)
\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)
b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)
c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)
\(B=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{51}\right)=\)\(\frac{250}{51}\)
\(B=5\left(\frac{1}{1}-\frac{1}{51}\right)=\frac{250}{51}\)
\(B=5\left(\frac{51}{51}-\frac{1}{51}\right)=\frac{250}{51}\)
\(B=5.\frac{50}{51}=\frac{250}{51}\)
\(B=\frac{250}{51}=\frac{250}{51}\)
Bài 1:
a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)
\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)
\(=\frac{-3}{5}\)
b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)
\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)
\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)
c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)
\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)
\(=3+2=5\)
d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)
\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)
\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)
e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)
\(=-1+1+\frac{-7}{9}\)
\(=-\frac{7}{9}\)
f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)
\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)
\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)
A=.....
=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY
TỰ LÀM NHE
\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)
\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
\(C=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)
\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)
\(C=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=9-\left(1-\frac{1}{10}\right)\)
\(C=9-\frac{9}{10}=\frac{81}{10}\)
A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)