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Có \(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{90}=\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{9.10}\)
= \(1-\frac{1}{2}-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{9}-\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}=\frac{9}{10}\)
kết quả là \(\frac{1}{10}\)nhà mình không biết cách làm vậy cho mình xin lỗi nha!
mình trả lời đầu tiên đó nha!
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
= - ( 1/2 +1/6+1/12+1/20+ 1/30+ 1/42+ 1/56+ 1/72+ 1/90)
= - ( 1 - 1/2 + 1/2 -1/3 +1/3 -1/4 +1/4 - 1/5 +1/5 -1/6 +1/6 -1/7 +1/7 -1/8 +1/8 -1/9 +1/9 -1/10)
= - ( 1- 1/10 )
= -9/10
\(-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\left(-\frac{1}{90}\right)+\left(-\frac{1}{72}\right)+\left(-\frac{1}{56}\right)+......+\left(-\frac{1}{2}\right)\)
\(=\left(-1\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{9.10}\right)\)
\(=\left(-1\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\left(-1\right).\left(1-\frac{1}{90}\right)=\frac{\left(-1\right).89}{90}=-\frac{89}{90}\)
nha
\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=1-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1-\left(1-\frac{1}{10}\right)\)
\(=1-\frac{9}{10}\)
\(=\frac{1}{10}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{2}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)(đáp án của p sai nha)
= 1 / 1*2 + 1 / 2*3 + 1/ 3*4 + 1 / 4 * 5 ... + 1/ 9*10 = 1-1/2 + 1/2 -1/3 +... + 1/9 - 1/10 = 1 - 1/10 = 9 /10
đáp án của bạn bị sai rùi
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 ~
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(A=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(A=\frac{9}{10}-\frac{9}{10}=0\)
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(\Leftrightarrow A=\frac{9}{10}-\frac{9}{10}\)
\(\Leftrightarrow A=0\)
\(S=\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)
\(S=\frac{3}{4}-\frac{1}{4}-\left[\frac{14}{6}+\left(\frac{-27}{6}\right)\right]-\frac{5}{6}\)
\(S=\frac{1}{2}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)
\(S=\frac{3}{6}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)
\(S=\frac{11}{6}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(\frac{9}{10}\)
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{30}\)+ ........ + \(\frac{1}{90}\)
= \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ \(\frac{1}{5.6}\)+ ....... + \(\frac{1}{9.10}\)
= \(\frac{2-1}{1.2}\)+ \(\frac{3-2}{2.3}\)+ \(\frac{4-3}{3.4}\)+ \(\frac{5-4}{4.5}\)+ \(\frac{6-5}{5.6}\)+ ......... + \(\frac{10-9}{9.10}\)
= \(\frac{2}{1.2}\)- \(\frac{1}{1.2}\)+ \(\frac{3}{2.3}\)- \(\frac{2}{2.3}\)+ \(\frac{4}{3.4}\)- \(\frac{3}{3.4}\)+ \(\frac{5}{4.5}\)- \(\frac{4}{4.5}\)+ \(\frac{6}{5.6}\)- \(\frac{5}{5.6}\)+ ........ + \(\frac{10}{9.10}\)- \(\frac{9}{9.10}\)
= 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{6}\)+ ........... + \(\frac{1}{9}\)- \(\frac{1}{10}\)
Sau đó ta trực tiêu:
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)