Question

\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)\(\frac{1}{2}\)

làm giúp mk với mk đang cần bài này

 

Answer 1

Ta có : \(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-.....-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{2}+\frac{1}{6}+.....+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}=\frac{1}{90}-\frac{80}{90}=\frac{-79}{90}\)

Answer 2

Đặt \(A=\left(...\right)\) ( tự ghi ) 

Ta có : 

\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(A=\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)

\(-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}-\frac{1}{9.10}\)

\(-A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(-\frac{1}{9}+\frac{1}{10}\)

\(-A=1-\frac{1}{9}-\frac{1}{9}+\frac{1}{10}\)

\(-A=\frac{79}{90}\)

\(A=\frac{-79}{90}\)

Chúc bạn học tốt ~ 

Answer 3

    \(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{10\cdot9}-\frac{1}{9\cdot8}-\frac{1}{8\cdot7}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)\

\(=\frac{1}{10}+\frac{1}{9}-\frac{1}{9}+\frac{1}{8}-\frac{1}{8}+\frac{1}{7}-...+\frac{1}{3}-\frac{1}{2}+1\)

\(=\frac{1}{10}+1\)

\(=\frac{11}{10}\)

                       Mình không chắc đâu nha bạn !

Answer 4

\(\text{Đ}\text{ặt} A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-....-\frac{1}{2}\)

\(A=\frac{1}{90}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}\right)\)

\(A=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\right)\)

\(A=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)\)

\(A=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(A=\frac{1}{90}-\frac{8}{9}\)

\(A=\frac{1}{90}-\frac{80}{90}\)

\(A=\frac{-79}{80}\)

Answer 5

\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{2}.\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)+\frac{1}{90}.\)

\(=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)

\(=-1+\frac{1}{9}+\frac{1}{90}\)

\(=\frac{-79}{90}\)