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Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
c,
= \(\dfrac{5}{9}.\left(\dfrac{7}{13}+\dfrac{9}{13}+\dfrac{-3}{13}\right)\)
= \(\dfrac{5}{9}.1\)
= \(\dfrac{5}{9}\)
1.
a) \(\dfrac{5}{18}+\dfrac{4}{7}+\dfrac{13}{18}+\dfrac{3}{7}\)
\(=\left(\dfrac{5}{18}+\dfrac{13}{18}\right)+\left(\dfrac{4}{7}+\dfrac{3}{7}\right)\)
\(=1+1=2\)
b) \(\dfrac{4}{9}.\dfrac{5}{19}.\dfrac{9}{4}\)
\(=\left(\dfrac{4}{9}.\dfrac{9}{4}\right).\dfrac{5}{19}\)
\(=1.\dfrac{5}{19}=\dfrac{5}{19}\)
tik mik nha!!!
2) \(\dfrac{4}{9}.\dfrac{5}{19}.\dfrac{9}{4} =(\dfrac{4}{9}.\dfrac{9}{4}).\dfrac{5}{19} =1.\dfrac{5}{19} =\dfrac{5}{19}\)
\(A=\dfrac{-9}{10^{2010}}+\dfrac{-19}{10^{2011}}=\dfrac{-90}{10^{2011}}+\dfrac{-19}{10^{2011}}=\dfrac{\left(-90\right)+\left(-19\right)}{10^{2011}}=\dfrac{-109}{10^{2011}}\)\(B=\dfrac{-9}{10^{2011}}+\dfrac{-19}{10^{2010}}=\dfrac{-9}{10^{2011}}+\dfrac{-190}{10^{2011}}=\dfrac{\left(-9\right)+\left(-190\right)}{10^{2011}}=\dfrac{-199}{10^{2011}}\)\(\text{Vì }\dfrac{-109}{10^{2011}}>\dfrac{-199}{10^{2011}}\text{ nên }A>B\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
câu a
\(\left(5\dfrac{7}{9}+2\dfrac{3}{11}\right)-\left(-1\dfrac{8}{11}+3\dfrac{7}{9}\right)\)
= \(5\dfrac{7}{9}+2\dfrac{3}{11}+1\dfrac{8}{11}-3\dfrac{7}{9}\)
= \(\left(5\dfrac{7}{9}-3\dfrac{7}{9}\right)+\left(2\dfrac{3}{11}+1\dfrac{8}{11}\right)\)
= \(2+3\dfrac{11}{11}\)
= \(2+3+1\)
= \(6\)
câu b
\(\left(2\dfrac{7}{19}-3\dfrac{8}{9}\right)-\left(1\dfrac{7}{19}-2\dfrac{1}{9}\right)\)
= \(2\dfrac{7}{19}-3\dfrac{8}{9}-1\dfrac{7}{19}+2\dfrac{1}{9}\)
= \(\left(2\dfrac{7}{19}-1\dfrac{7}{19}\right)-\left(3\dfrac{8}{9}-2\dfrac{1}{9}\right)\)
= \(1-1\dfrac{7}{9}\)
= \(\dfrac{-7}{9}\)
\(A=15.\left(\dfrac{3}{5}-\dfrac{2}{3}\right)+1\\ A=15.\left(\dfrac{9}{15}-\dfrac{10}{15}\right)+1\\ A=15.\dfrac{-1}{15}+1\\ A=-1+1\\ A=0\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\\ C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{9}.\dfrac{9}{11}+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.1+\dfrac{12}{7}\\ C=\dfrac{-5}{7}+\dfrac{12}{7}\\ C=1\)
\(\left(a\right):P=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{99}{100}\)
Nhận xét
thừa số tổng quát là \(\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\) với n =1 đến 10
\(P=\dfrac{1.3.2.4.3.5...9.11}{2^2.3^2...9^2.10^2}=\dfrac{\left(1.2.3...9\right)\left(3.4.5....11\right)}{\left(2.3.4....10\right)\left(2.3.4....10\right)}\)
\(P=\dfrac{1.2.3..9}{2.3.4..9.10}.\dfrac{3.4.5...10.11}{2.3.4....10}=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)
\(C=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2^2\right)\\ C=\dfrac{6}{7}+\dfrac{5}{8}.\dfrac{1}{5}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\dfrac{-4}{1}\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{-3}{4}\\ C=\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{-42}{56}\\ C=\dfrac{97}{56}\)
Lưu An
\(A=\dfrac{1}{19}+\left(\dfrac{9}{19\cdot29}+\dfrac{9}{29\cdot39}+...+\dfrac{9}{1999\cdot2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19\cdot29}+\dfrac{10}{29\cdot39}+...+\dfrac{10}{1999\cdot2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+...+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\cdot\dfrac{1990}{38171}\)
\(A=\dfrac{1}{19}+\dfrac{1791}{38171}\)
\(A=\dfrac{200}{2009}\)
B=1/19+(9/19.29+9/29.39+...+9/1999.2009)
B=1/19+9/10+(10/19.29+10/29.39+.....+10/1999.2009
B=1/19+9/10+(1/19-1/29+1/29-1/39+....+1/1999-1/2009)
B=1/19+9/10+(1/19-1/2009)
B=1/19+9/10.1990/38171
B=1/19+1791/38171
B=200/2009
Vậy B= 200/2009