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\(\dfrac{48}{25}\cdot\dfrac{27}{55}+2\dfrac{4}{9}\cdot\dfrac{14}{33}\)
\(=\dfrac{1296}{1375}+\dfrac{22}{9}\cdot\dfrac{14}{33}\\ =\dfrac{1296}{1375}+\dfrac{28}{27}\\ =\dfrac{34992}{37125}+\dfrac{38500}{37125}\\ =\dfrac{73492}{37125}\)
\(1\dfrac{19}{22}\cdot\left(\dfrac{47}{77}-\dfrac{16}{15}\right)\\ =\dfrac{41}{22}\cdot\dfrac{-527}{1155}\\ =\dfrac{-21607}{25410}\)
\(\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)-\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{307}{99}+\dfrac{37}{99}-\dfrac{1503}{299}\right)-0\\ =\dfrac{344}{99}-\dfrac{1053}{299}\\ =-\dfrac{107}{2277}\)
1:-5/14=-50/140
3/20=21/140
9/70=18/140
2: -55/132=-5/12=-35/84
10/42=20/84
-3/28=-9/84
3: 7/10=231/330
1/33=10/330
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)
\(CóS=\dfrac{3}{2^2}+\dfrac{8}{3^2}+\dfrac{15}{4^2}+...+\dfrac{2499}{50^2}\)
\(Tathấy\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=1-\dfrac{1}{50}< 1\)\(-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>-1\)
\(S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1\)\(S=\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>48\)
\(\rightarrow S>48\)
\(VậyS>48\)
a: -9/20=-36/320
17/320=17/320
b: -7/10=-231/330
1/33=10/330
c: -5/15=-1/3=-140/420
3/20=63/420
9/70=54/420
d: 10/42=5/21=20/84
-3/28=-9/84
-55/132=-5/12=-35/84
\(\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\cdot\cdot\cdot\left(1+\dfrac{7}{180}\right)=\dfrac{16}{9}\cdot\dfrac{27}{20}\cdot\cdot\cdot\dfrac{187}{180}=\dfrac{2.8}{1\cdot9}\cdot\dfrac{3\cdot9}{2\cdot10}\cdot\cdot\cdot\dfrac{11\cdot17}{10\cdot18}=\dfrac{\left(2\cdot3\cdot...\cdot11\right)\cdot\left(8\cdot9\cdot...\cdot17\right)}{\left(1\cdot2\cdot...\cdot10\right)\cdot\left(9\cdot10\cdot...\cdot18\right)}=\dfrac{11\cdot8}{1\cdot18}=\dfrac{88}{18}=\dfrac{44}{9}\)
Ta có:
\(S=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}...\dfrac{42.48}{41.49}\)
\(S=\dfrac{2.3.4...42}{1.2.3...41}.\dfrac{8.9.10...48}{9.10.11...49}\)
\(S=\dfrac{42}{1}.\dfrac{8}{49}\)
\(S=\dfrac{48}{7}\)
Giải:
\(S=\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}.\dfrac{55}{48}.....\dfrac{2016}{2009}.\)
\(=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}.\dfrac{5.11}{4.12}.....\dfrac{42.48}{41.49}.\)
\(=\dfrac{\left(2.3.4.5.....42\right)\left(8.9.10.11.....48\right)}{\left(1.2.3.4.....41\right)\left(9.10.11.12.....49\right)}.\)
\(=\dfrac{2.3.4.5.....42}{1.2.3.4.....41}.\dfrac{8.9.10.11.....48}{9.10.11.12.....49}.\)
\(=42.\dfrac{8}{49}.\)
\(=\dfrac{42.8}{49}.\)
\(=\dfrac{6.8}{7}.\)
\(=\dfrac{48}{7}.\)
Vậy \(S=\dfrac{48}{7}.\)
CHÚC BN HỌC TỐT!!! ^ - ^
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