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a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
=>(y-1/2):(1-1/2+1/2-1/3+...+1/9-1/10)=1/6
=>(y-1/2):9/10=1/6
=>(y-1/2)=1/6*9/10=9/60=3/20
=>y=3/20+10/20=13/20
\(A=\dfrac{1}{3}\cdot\dfrac{4}{5}+\dfrac{1}{3}\cdot\dfrac{6}{5}+\dfrac{2}{3}\\ =\dfrac{1}{3}\cdot\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\\ =\dfrac{1\cdot2}{3}+\dfrac{2}{3}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{4}{3}\)
B =\(\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}\)-\(\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\left[\dfrac{-5}{6}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\left[\dfrac{-10}{12}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)
=\(\dfrac{-17}{57}\)-\(\dfrac{40}{57}\)
=\(\dfrac{-17}{57}+\dfrac{-40}{57}\)
=\(\dfrac{-57}{57}=-1\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
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!!! ~
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-1-\dfrac{1}{2}-...-\dfrac{1}{10}\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
Vậy \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
a, \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}\)
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{100}\)
\(\Rightarrow\dfrac{99}{100}\)
thế phần b ko làm à ???
là quên hay ko biết làm