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Mình giải theo kiểu lớp 6 nhá !
=\(33.\left(\frac{34}{15}+\frac{34}{35}+\frac{34}{63}+\frac{34}{99}\right)\)
=\(33.\left[34.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]\)
=\(33.\left[34.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]\)
=\(33.\left[34.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).\frac{1}{2}\right]\)
=\(33.\left[34\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).\frac{1}{2}\right]\)
=\(33.\left[34.\left(\frac{1}{3}-\frac{1}{11}\right).\frac{1}{2}\right]\)
=\(33.\left(34.\frac{8}{33}.\frac{1}{2}\right)\)
=\(33.\frac{136}{33}\)
=\(\frac{33.136}{33}\)(*)
=\(136\)
(Bạn có thể bỏ bước có dấu *)
\(\frac{1313}{1414}=\frac{13.101}{14.101}=\frac{13}{14}\)
=) \(\frac{13}{14}=\frac{13}{14}\)
\(\frac{2222}{3434}=\frac{22.101}{34.101}=\frac{22}{34}\)
\(\frac{12012}{17017}=\frac{12.1001}{17.1001}=\frac{12}{17}\)
Ta có : \(\frac{22}{34}\)và\(\frac{12}{17}\)
=) \(\frac{22}{34}\)và \(\frac{24}{34}\)
Phan số bên trái có tử số lớn hơn nên \(\frac{24}{34}\)>\(\frac{22}{34}\)
Đăng kí kênh của V-I-S nha !
Giải
Bài A) \(\frac{191919}{212121}\)+ \(\frac{88888}{99999}\)= \(\frac{133}{63}\)
Bài B) \(\frac{3434}{3535}\)+ \(\frac{343434}{6565650}\)= \(\frac{2329}{2275}\)^0^
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
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1\cdot2\cdot3\cdot4}{2\cdot3\cdot4\cdot5}=\frac{1}{5}\)
\(\frac{7}{4}\)x(\(\frac{33}{12}\)+\(\frac{33}{20}\)+\(\frac{33}{30}\)+\(\frac{33}{42}\))
=\(\frac{7}{4}\)x\(\frac{44}{7}\)
=\(\frac{308}{28}\)=11
=4/5x5/6x...x99/100
=4x5x6x...x99/5x6x7x...x100
=4/100=1/25
( 3434/3535 + 3434/6363 + 3434/9999 ) × 55
= ( 34/35 + 34/63 + 34/99 ) × 55
= 34 × ( 1/35 + 1/63 + 1/99 ) × 55
= 34 × 3/55 × 55
= 102/55 × 55
= 102
Mình ko chắc lắm đâu nhá