Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-.....-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{2}+\frac{1}{6}+.....+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{1}{90}-\frac{80}{90}=\frac{-79}{90}\)
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 ~
a) 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/10.9 - 1/9.8 - 1/8.7 - 1/7.6 - 1/6.5 - 1/5.4 - 1/4.3 - 1/3.2 - 1/2.1
= 1/10 - 1
= 0,1 - 1
= -0,9
Ta có :
\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=2017.\frac{1}{2017}=1\)
\(\Rightarrow A=1-3=-2\)
a) \(-5,13:\left(5\frac{5}{28}-1\frac{8}{9}.1,25+1\frac{16}{63}\right)\)
\(=-5,13:\left(\frac{145}{28}-\frac{17}{9}.1,25+\frac{79}{63}\right)\)
\(=-5,13:\left(\frac{145}{28}-\frac{85}{36}+\frac{79}{63}\right)\)
\(=-5,13:\frac{57}{14}\)
\(=-\frac{63}{50}\)
b) \(\left(3\frac{1}{3}.1,9+19,5:4\frac{1}{3}.\left(\frac{62}{75}-\frac{4}{25}\right)\right)\)
\(=\left(\frac{10}{3}.1,9+19,5:\frac{13}{3}\right).\left(\frac{62}{75}-\frac{4}{25}\right)\)
\(=\left(\frac{19}{3}+\frac{9}{2}\right).\frac{2}{3}\)
\(=\frac{65}{6}.\frac{2}{3}\)
\(=\frac{65}{9}\)
^...^ 
^_^
\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
\(-A=\left(\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(-A=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+\frac{1}{8}-\frac{1}{7}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)
\(-A=\frac{1}{10}-1=\frac{-9}{10}\Rightarrow A=\frac{9}{10}\)
\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
Vậy A=-79/90