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a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)
\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)
\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)
\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)
\(\left(a\right):P=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{99}{100}\)
Nhận xét
thừa số tổng quát là \(\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\) với n =1 đến 10
\(P=\dfrac{1.3.2.4.3.5...9.11}{2^2.3^2...9^2.10^2}=\dfrac{\left(1.2.3...9\right)\left(3.4.5....11\right)}{\left(2.3.4....10\right)\left(2.3.4....10\right)}\)
\(P=\dfrac{1.2.3..9}{2.3.4..9.10}.\dfrac{3.4.5...10.11}{2.3.4....10}=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{899}{900}\)
\(=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times...\times\frac{29.31}{30.30}\)
\(=\frac{\left(1\times2\times3\times...\times29\right)\left(3\times4\times5\times...\times31\right)}{\left(2\times3\times4\times...\times30\right)\left(2\times3\times4\times...\times30\right)}\)
\(=\frac{1\times2\times3\times...\times29}{2\times3\times4\times...\times30}.\frac{3\times4\times5\times...\times31}{2\times3\times4\times...\times30}\)
\(=\frac{1}{30}.\frac{31}{2}\)
\(=\frac{31}{60}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{899}{900}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{29.31}{30.30}\\ =\frac{1.2.3.4....29}{2.3.4...30}.\frac{3.4.5...31}{2.3.4...30}\\ =\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
.
\(a.\)
\(-\dfrac{5}{9}\cdot\dfrac{12}{35}=\dfrac{\left(-5\right)\cdot12}{9\cdot35}=\dfrac{-60}{315}=-\dfrac{4}{21}\)
\(b.\)
\(\left(-\dfrac{5}{8}\right)\cdot-\dfrac{6}{55}=\dfrac{\left(-5\right)\cdot\left(-6\right)}{8\cdot55}=\dfrac{30}{440}=\dfrac{3}{44}\)
\(c.\)
\(\left(-7\right)\cdot\dfrac{2}{5}=-\dfrac{14}{5}\)
\(d.\)
\(-\dfrac{3}{8}\cdot\left(-6\right)=\dfrac{-3\cdot\left(-6\right)}{8}=\dfrac{18}{8}=\dfrac{9}{4}\)
Bài 1:
\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)
\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)
\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)
a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)
Ta có: \(B=\dfrac{5}{13}\cdot\dfrac{8}{15}+\dfrac{5}{13}\cdot\dfrac{26}{15}-\dfrac{5}{13}\cdot\dfrac{8}{15}\)
\(=\dfrac{5}{13}\cdot\dfrac{26}{15}\)
\(=\dfrac{130}{195}=\dfrac{2}{3}\)
\(=\dfrac{1.4.9.16}{2.6.12.20}=\dfrac{576}{2880}=\dfrac{1}{5}\)
\(\dfrac{1}{2}\cdot\dfrac{4}{6}\cdot\dfrac{9}{12}\cdot\dfrac{16}{20}=\dfrac{1}{2}\cdot\dfrac{2\cdot2}{2\cdot3}\cdot\dfrac{3\cdot3}{3\cdot4}\cdot\dfrac{4\cdot4}{4.5}=\dfrac{1}{5}\)
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\\ A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\\ A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot29\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot31\right)}{\left(2\cdot3\cdot4\cdot...\cdot30\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot30\right)}\\ A=\dfrac{1\cdot2\cdot3\cdot...\cdot29}{2\cdot3\cdot4\cdot...\cdot30}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot31}{2\cdot3\cdot4\cdot...\cdot30}\\ A=\dfrac{1}{30}\cdot\dfrac{31}{2}\\ A=\dfrac{31}{60}\)
A=\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{899}{900}\)
A=\(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
A=\(\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
A=\(\dfrac{\left(1.2.3.....28.29\right).\left(3.4.5....29.31\right)}{\left(2.3.4....29.30\right).\left(2.3.4....29.30\right)}\)
A=\(\dfrac{1.31}{30.2}\)
A=\(\dfrac{31}{60}\)
Vậy A = \(\dfrac{31}{60}\)