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100 - 99 + 98 - 97 + 96 - 95 +..... + 4 - 3 + 2
= (100 - 99 ) + ( 98 - 97 ) + ( 96 - 95 ) + ..... + (4 - 3 ) + 2
= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + ...... + 1 + 1 + 2
= .......... ( bn tự tính nhé )
Hok tốt ! ^_^
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
100 + 99 - 98 - 97 + 96 + 95 - 94 - 93 + .......... + 4 + 3 - 2 - 1
= ( 100 + 99 - 98 - 97 ) + ( 96 + 95 - 94 - 93 ) + ......... + ( 4 + 3 - 2 - 1 ) ( 25 nhóm )
= 4 + 4 .......+ 4 ( 25 số 4 )
= 25 . 4
= 100
làm :
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{3}{8}\)
b, \(ab\cdot10-ab=2ab\)
\(ab\cdot10-ab\cdot1=2ab\)
\(ab\cdot\left(10-1\right)=2ab\)
\(ab\cdot9=2ab\)
\(ab\cdot9=200+ab\cdot1\)
\(ab\cdot9-ab\cdot1=200\)
\(ab\cdot\left(9-1\right)=200\)
\(ab\cdot8=200\)
\(ab=200:8\)
\(ab=25\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{101}-\frac{1}{103}\)
\(A=\frac{1}{3}-\frac{1}{103}\)
\(A=\frac{100}{309}\)
\(A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}+\frac{2}{101\times103}\)
\(A=1\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}+\frac{1}{101}-\frac{1}{103}\right)\)
\(A=1\times\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(A=1\times\frac{100}{309}\)
\(A=\frac{100}{309}\)
\(100-99+98-97+96-95+...+4-3+2\)
\(=(100-99)+(98-97)+(96-95)+...+(4-3)+2\)
\(=1+1+1+...+1+2\)
\(=98+2=100\)
P/S : Đúng ko ta ???
Bài làm
= \(\frac{18.25+9.45.2+3.27.6}{100-99+98-97+96-95+.....+2-1}\)
= \(\frac{18.25+9.2.45+3.6.27}{50}\)
= \(\frac{18.25+18.45+18.27}{50}\)
= \(\frac{18.\left(25+45+27\right)}{50}\)
= \(\frac{18.97}{50}\)
= \(\frac{1746}{50}\)
50 lấy đâu ra, phải tính ra thể thống chứ