1.tính thuận tiện
a) 1+ \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)
b) 45 \(\times\) 48 - 90 \(\times\) 24 + 145
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a) \(\dfrac{2}{3}\times\dfrac{1}{4}-\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{2}{12}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=\dfrac{0}{6}=0\)
b) \(\dfrac{8}{5}\times\dfrac{1}{4}-\dfrac{2}{5}\times\dfrac{1}{2}-\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{8}{20}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4-2-1}{10}=\dfrac{1}{10}\)
\(\dfrac{15}{16}:\dfrac{5}{8}\times\dfrac{3}{4}\)
\(=\dfrac{15}{16}\times\dfrac{8}{5}\times\dfrac{3}{4}\)
\(=\dfrac{3}{2}\times\dfrac{3}{4}\)
\(=\dfrac{9}{8}\)
_________________
\(\dfrac{21}{4}\times\dfrac{16}{14}\times\dfrac{1}{2}\times\dfrac{8}{3}\)
\(=6\times\dfrac{4}{3}\)
\(=8\)
a) $\frac{2}{5} \times \frac{3}{8} \times \frac{3}{4} = \frac{{2 \times 3 \times 3}}{{5 \times 8 \times 4}} = \frac{{18}}{{160}} = \frac{9}{{80}}$
b) $\frac{1}{3} \times \frac{1}{6} \times \frac{1}{9} = \frac{{1 \times 1 \times 1}}{{3 \times 6 \times 9}} = \frac{1}{{162}}$
c) $\frac{3}{4}:\frac{1}{5}:\frac{7}{8} = \frac{3}{4} \times \frac{5}{1} \times \frac{8}{7} = \frac{{3 \times 5 \times 8}}{{4 \times 1 \times 7}} = \frac{{120}}{{28}} = \frac{{30}}{7}$
d) $\frac{3}{5}:\frac{1}{5}:\frac{3}{8} = \frac{3}{5} \times \frac{5}{1} \times \frac{8}{3} = \frac{{3 \times 5 \times 8}}{{5 \times 1 \times 3}} = 8$
a: \(=5-2\cdot\dfrac{1}{4}=5-\dfrac{1}{2}=\dfrac{9}{2}\)
b: \(=\left(\dfrac{7}{2}\right)^3+\dfrac{1}{2}=\dfrac{343}{8}+\dfrac{1}{2}=\dfrac{347}{8}\)
c: \(=\left(5+\dfrac{5}{27}-\dfrac{5}{27}\right)+\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\dfrac{1}{2}=5+1-\dfrac{1}{2}=5+\dfrac{1}{2}=5.5\)
e: \(=\dfrac{-5}{4}\left(35+\dfrac{1}{6}-45-\dfrac{1}{6}\right)=\dfrac{-5}{4}\cdot\left(-10\right)=\dfrac{50}{4}=\dfrac{25}{2}\)
\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)
\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)
\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)
\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)
\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)
\(A\cdot\left(1-\dfrac{1}{4}\right)\cdot\left(1-\dfrac{1}{9}\right)\cdot\left(1-\dfrac{1}{16}\right)\left(1-\dfrac{1}{25}\right)=1\dfrac{3}{5}\)
=>\(A\cdot\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{5}\right)=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{5}\cdot\dfrac{6}{2}=\dfrac{8}{5}\)
=>\(A\cdot3=8\)
=>A=8/3
a: =4/5+1/5+2/3+1/3=1+1=2
b: =17/12+7/12+29/7-8/7=3+2=5
c: =3/5+2/5+16/7-1/7-1/7
=1+2=3
d: =2/5+3/5+2/3+1/3+7/4+1/4
=1+1+2
=4
\(a)N=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(2N=2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}\)
\(2N-N=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\right)\)
\(N=2-\dfrac{1}{16}\)
\(N=\dfrac{31}{16}\)
\(b)45\times48-90\times24+145\)
\(=45\times48-45\times48+145\)
\(=0+145\)
\(=145\)