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\(A=\frac{8\cdot4\cdot125\cdot25+96524+2476}{10\cdot125\cdot4\cdot25\cdot8}\)
\(A=\frac{1+96524+2476}{10}\)
\(A=\frac{99001}{10}\)
\(B=\frac{5+55+555+5555}{9+99+999+9999}\)
\(B=\frac{9}{5}\)
1 1/9 x 1 1/10 x 1 1/11 x ... x 1 1/2011
=10/9 x 11/10 x 12/11 x ... x 2012/2011
khử
còn 2012/9
=\(\frac{10}{9}\)x\(\frac{11}{10}\)x\(\frac{12}{11}\)x.........x\(\frac{2012}{2011}\)
=\(\frac{2012}{9}\)
1/1x2+1/2x3+1/3x4+...+1/99x100+1
1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100 +1
=1- 1/100 +1
=99/100 +1
=199/100
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/999.1000 + 1
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/999 - 1/1000 + 1
= 1 - 1/1000 + 1
= 2 - 1/1000
= 2000/1000 - 1/1000
= 1999/1000
Ủng hộ mk nha ♡_♡☆_☆
Gọi \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=1-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=\frac{24}{25}\)
\(\Leftrightarrow\)\(A=\frac{24}{25}:3\)
\(\Leftrightarrow\)\(A=\frac{24}{25}.\frac{1}{3}\)
\(\Leftrightarrow\)\(A=\frac{8}{25}\)
Vậy \(A=\frac{8}{25}\)
Đặt \(C=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}\)
\(\Rightarrow3C=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{22.25}\)
\(\Rightarrow3C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Rightarrow3C=1-\frac{1}{25}=\frac{24}{25}\)
\(\Rightarrow C=\frac{24}{25}:3=\frac{8}{25}\)
Vậy \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}=\frac{8}{24}\)
\(C=\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.8}+\frac{1}{9.10}\right)\)
\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(C=\frac{9}{10}-\frac{9}{10}=0\)
\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}.\)
\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)
\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)
\(\Leftrightarrow x\cdot\frac{5}{3}=15\)
\(\Leftrightarrow x=15:\frac{5}{3}\)
\(\Leftrightarrow x=15\cdot\frac{3}{5}\)
\(\Leftrightarrow x=9.\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)
\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)
\(\Rightarrow x.\frac{5}{3}=14+1=15\)
\(\Rightarrow x=15:\frac{5}{3}=9\)
Bùi Thùy Linh
21/20 nha bam
k tui nha
thank
\(\frac{63.303}{505.36}=\frac{9.7.101.3}{101.5.9.4}=\frac{21}{20}\)