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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right).200x=4036\)
\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}.200x=4036\)
\(\Leftrightarrow\frac{1.2.3...99}{2.3.4....100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow2x=4036\)
\(\Leftrightarrow x=4036:2=2018\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)
=> \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}\times200\times x=4036\)
=> \(\frac{1\times2\times...\times99}{2\times3\times...\times100}\times200\times x=4036\)
\(\Rightarrow\frac{1}{100}\times200\times x=4036\)
\(\Rightarrow2\times x=4036\)
=> x = 2018
\(a,\frac{1}{2}\times\frac{4}{5}\div\frac{6}{10}\)
\(=\frac{2}{5}\div\frac{6}{10}\)
\(=\frac{2}{5}\times\frac{10}{6}\)
\(=\frac{2}{3}\)
\(b,\frac{24}{35}\div\left(\frac{4}{5}\times\frac{8}{7}\right)\)
\(=\frac{24}{35}\div\frac{32}{35}\)
\(=\frac{24}{35}\times\frac{35}{32}\)
\(=\frac{3}{4}\)
bài 1
Ta có : 2016/2017<1
2017/2018<1
Nên 2016/2017=2017/2018
Bài 1 :
a) Ta có : \(\frac{2016}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=1-\frac{1}{2018}\)
Vì \(-\frac{1}{2017}< -\frac{1}{2018}\)nên \(\frac{2016}{2017}< \frac{2017}{2018}\)
b) Ta có : \(\frac{2018}{2017}=1+\frac{1}{2017}\)
\(\frac{2017}{2016}=1+\frac{1}{2016}\)
Vì \(\frac{1}{2017}< \frac{1}{2016}\) nên \(\frac{2018}{2017}< \frac{2017}{2016}\)
Câu 2 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
bài 1:
\(\frac{6}{11}+\frac{1}{3}+\frac{5}{11}\)
\(=\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{1}{3}\)
\(=\frac{11}{11}+\frac{1}{3}=1+\frac{1}{3}=\frac{3}{3}+\frac{1}{3}=\frac{4}{3}\)
bài 2:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{12}\right)\)
\(=\frac{11}{20}+\frac{1}{4}=\frac{11}{20}+\frac{5}{20}=\frac{15}{20}=\frac{3}{4}\)
bài 3:
a) \(\frac{3}{2}\cdot\frac{4}{5}\cdot\frac{2}{3}=\left(\frac{3}{2}\cdot\frac{2}{3}\right)\cdot\frac{4}{5}=1\cdot\frac{4}{5}=\frac{4}{5}\)
b) \(\frac{6}{7}\cdot\frac{5}{3}\cdot\frac{7}{6}=\left(\frac{6}{7}\cdot\frac{7}{6}\right)\cdot\frac{5}{3}=1\cdot\frac{5}{3}=\frac{5}{3}\)
bài 4:
a) \(\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{4}\cdot\frac{2}{5}=\frac{2}{5}\cdot\left(\frac{1}{4}+\frac{3}{4}\right)=\frac{2}{5}\cdot1=\frac{2}{5}\)
b) \(\frac{6}{11}:\frac{2}{3}+\frac{5}{11}:\frac{2}{3}=\left(\frac{6}{11}+\frac{5}{11}\right):\frac{2}{3}=1:\frac{2}{3}=\frac{3}{2}\)
Bài 1:
6/11 + 1/3 + 5/11
= ( 6/11 + 5/11) + 1/3
= 11/11 + 1/3
= 1 + 1/3
= 3/3 +1/3
= 4/3
Bài 2:
1/2 + 1/6 + 1/12 + 1/20
= ( 1/2 + 1/6 + 1/12 ) + 1/20
= ( 6/12 + 2/12 + 1/12 ) + 1/20
=9/12 + 1/20
= 3/4 +1/20
= 15/20 + 1/20
= 16/20 = 4/5
Bài 3:
a) \(\frac{3}{2}\times\frac{4}{5}\times\frac{2}{3}\) \(=\left(\frac{3}{2}\times\frac{2}{3}\right)\times\frac{4}{5}\)\(=1\times\frac{4}{5}=\frac{4}{5}\)
b) \(\frac{6}{7}\times\left(\frac{5}{3}\times\frac{7}{6}\right)\) \(=\frac{6}{7}\times\frac{35}{18}\)\(=\frac{1\times5}{7\times3}=\frac{5}{21}\)
Bài 4:
a) 2/5 x 1/4 + 3/4 x 2/5
= 2/5 x ( 1/4 + 3/4)
= 2/5 x 1
= 2/5
b) 6/11 : 2/3 +5/11 : 2/3
= ( 6/11 + 5/11) x 3/2
= 11/11 x 3/2
= 1 x 3/2
= 3/2
....
1. \(\frac{14}{45}=\frac{1}{9}+\frac{1}{5}\)
2. \(\left(1-\frac{1}{12}\right).\left(1-\frac{1}{11}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{8}\right)\)
\(=\frac{11}{12}.\frac{10}{11}.\frac{9}{10}.\frac{8}{9}.\frac{7}{8}\)
Triệt tử với mẫu:
\(=\frac{7}{12}\)
1.ket qua la 1/5+1/9
2.=11/12x10/11x9/10x8/9x7/8
=(11x10x9x8x7)/(12x11x10x9x8)
=7/12
câu 2:
A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/98*99 + 1/99*100
A = 2-1/1*2 + 3-2/2*3 + 4-3/3*4 + ... + 99-98/98*99 + 100-99/99*100
A = 2/1*2 - 1/ 1*2 + 3/2*3 - 2/2*3 + 4/ 3*4 -3/3*4 +...+ 99/98*99 - 98/98*99 + 100/99*100 - 99/99*100
A = 1 - 1/ 100
A = 99 / 100
phần 2 mk ko =bít
bài 1, 3 mk ko bít
1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko