TÍNH HỢP LÍ
A = \(\frac{-1}{20}\)+ \(\frac{-1}{30}\)+ \(\frac{-1}{42}\)+ \(\frac{-1}{56}\)+ \(\frac{-1}{72}\)+ \(\frac{-1}{90}\)
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23 tháng 1 2020
\(a,A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)
\(=\frac{-1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}-...-\frac{1}{9}+\frac{1}{10}\)
\(=-\frac{1}{4}+\frac{1}{10}\)
\(=-\frac{3}{20}\)
\(b,B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{11.7}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-....-\frac{1}{28}\)
\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
CC
23 tháng 1 2020
a) \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(\Rightarrow-1.A=\frac{1}{20}+\frac{1}{30}+........+\frac{1}{90}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+........+\frac{1}{9.10}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{9}-\frac{1}{10}=\frac{1}{4}-\frac{1}{10}=\frac{3}{20}\)
\(\Rightarrow A=\frac{3}{20}:\left(-1\right)=\frac{-3}{20}\)
b) \(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\Rightarrow\frac{1}{7}B=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(\Rightarrow B=\frac{13}{28}:\frac{1}{7}=\frac{13}{28}.7=\frac{13}{4}\)
8 tháng 8 2016
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
XO
17 tháng 9 2020
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)
NN
17 tháng 9 2020
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+.....+\frac{1}{2}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+........+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)
XO
6 tháng 7 2019
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}...-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
1 tháng 9 2019
"girl cute" là sai rồi bạn ơi, trong tiếng anh, tính từ (cute) phải đứng trước danh tư (girl).
19 tháng 4 2016
\(A=-\frac{1}{20}+-\frac{1}{30}+...+-\frac{1}{90}\)
\(=-\frac{1}{4.5}+-\frac{1}{5.6}+...+-\frac{1}{9.10}\)
\(=\left(-\frac{1}{4}\right)-\left(-\frac{1}{5}\right)+\left(-\frac{1}{5}\right)-\left(-\frac{1}{6}\right)+...+\left(-\frac{1}{9}\right)-\left(-\frac{1}{10}\right)\)
\(=\left(-\frac{1}{4}\right)-\left(-\frac{1}{10}\right)=-\frac{3}{20}\)
Vậy \(A=-\frac{3}{20}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
28 tháng 8 2020
\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3}-\frac{1}{9}\right)=1+\frac{1}{2}-\frac{1}{3}=1\frac{1}{6}\)
19 tháng 7 2018
\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+...+-\frac{1}{90}\)
\(\Leftrightarrow A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(A=-\frac{3}{20}\)
27 tháng 3 2017
\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90} \)\(\frac{-1}{90}\)
\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)
\(=\frac{-1}{4}-\frac{-1}{5}+\frac{-1}{5}-\frac{-1}{6}+\frac{-1}{6}-\frac{-1}{7}+\frac{-1}{7}-\frac{-1}{8}+\frac{-1}{8}-\frac{-1}{9}+\frac{-1}{9}-\frac{-1}{10}\)
\(=\frac{-1}{4}-\frac{-1}{10} \)\(=\frac{-3}{20}\)
PV
27 tháng 3 2017
\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
= \(\frac{-1}{4}-\frac{-1}{10}\)
=\(\frac{-3}{20}\)
T
12 tháng 4 2019
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{2}{5}\)
H
12 tháng 4 2019
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow A=\frac{2}{5}\)
BẰNG \(\frac{-3}{20}\)
A=\(-1\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}.+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
=\(-1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
=\(-1\left(\frac{1}{4}-\frac{1}{10}\right)\)=\(-1.\frac{3}{20}=\frac{-3}{20}\)