Rút gọn phan sô:
\(\frac{1339}{1442}\)và \(\frac{1212}{3030}\)
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\(\frac{1212}{3030}=\frac{2}{5}\)
\(\frac{32032}{48048}=\frac{2}{3}\)
\(\frac{456456}{234234}=\frac{76}{39}\)
\(\frac{1339}{1442}=\frac{13}{14}\)
#Châu's ngốc
1) Ta có : \(\frac{49}{28}=\frac{49:7}{28:7}=\frac{7}{4}\)
\(\frac{85}{51}=\frac{85:17}{51:17}=\frac{5}{3}\)
\(\frac{64}{96}=\frac{64:32}{96:32}=\frac{2}{3}\)
\(\frac{1212}{3030}=\frac{1212:606}{3030:606}=\frac{2}{5}\)
\(\frac{32032}{48048}=\frac{32032:16016}{48048:16016}=\frac{2}{3}\)
\(\frac{1339}{1442}=\frac{1339:103}{1442:103}=\frac{13}{14}\)
Bài 1 :
49/28 =7/4 85/51=5/3 64/96=2/3 1212/3030=2/5 32032/48048=2/3 1339/1442=13/14
\(\dfrac{1339}{1442}=\dfrac{1339:103}{1442:103}=\dfrac{13}{14}\)
\(\dfrac{1212}{3030}=\dfrac{1212:606}{3030:606}=\dfrac{2}{5}\)
A=\(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
A=\(\frac{7}{4}.\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)
A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)
A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3-4}+\frac{1}{4-5}+\frac{1}{5-6}+\frac{1}{6-7}\right)\right]\)
A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
A=\(\frac{7}{4}.\frac{44}{7}\)
A=11
Like cho mình nha bài này viết mỏi tay lắm
\(D=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
\(D=\frac{2002.101}{101.12}+\frac{2002.101}{20.101}+\frac{2002.101}{30.101}+\frac{2002.101}{42.101}+\frac{2002.101}{56.101}\)
\(D=\frac{2002}{12}+\frac{2002}{20}+\frac{2002}{30}+\frac{2002}{42}+\frac{2002}{56}\)
\(D=\frac{1001}{6}+\frac{1001}{10}+\frac{1001}{15}+\frac{143}{3}+\frac{143}{4}\)
\(D=\frac{5005}{12}\)
A = 7/4 . (3333/1212 + 3333/2020 + 3333/3030 + 3333/4242)
A = 7/4 . (11/4 + 33/20 + 11/10 + 11/14)
A = 7/4 . 44/7
A = 11
Chúc bạn học tốt
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{42.101}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{6}\right)\)
\(A=\frac{7}{4}.33.\frac{1}{6}\)
\(A=\frac{7.33}{4.6}\)
\(A=\frac{7.3.11}{4.3.2}\)
\(A=\frac{7.11}{4.2}\)
\(A=\frac{77}{8}\)
\(\frac{1339}{1442}=\frac{13}{14}\)
\(\frac{1212}{3030}=\frac{2}{5}\)
\(\frac{1339}{1442}=\frac{1300+39}{1400+42}=\frac{13\times100+13\times3}{14\times100+14\times3}=\frac{13\times\left(100+3\right)}{14\times\left(100+3\right)}=\frac{13}{14}\)\(\frac{13}{14}\)
\(\frac{1212}{3030}=\frac{1200+12}{3000+30}=\frac{12\times100+12}{30\times100+30}=\frac{12\times\left(100+1\right)}{30\times\left(100+1\right)}=\frac{12}{30}=\frac{2}{5}\)