Tính:
A= 2 + 2^2 + 2^3 + 2^4 + 2^5 +2^6 + 2^7 + 2^8
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a) \(\dfrac{6}{7}+\dfrac{7}{8}=\dfrac{48}{56}+\dfrac{49}{56}=\dfrac{97}{56}\)
b) \(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\)
c) \(\dfrac{2}{3}.\dfrac{4}{9}=\dfrac{8}{27}\)
d) \(\dfrac{1}{5}:\dfrac{2}{7}=\dfrac{1}{5}.\dfrac{7}{2}=\dfrac{7}{10}\)
a: =48/56+49/56
=97/56
b: =12/15-10/15
=2/15
c: =(2*4)/(3*9)=8/27
d: =1/5*7/2=7/10
a: \(=\dfrac{6}{7}\cdot\dfrac{-3}{5}=\dfrac{-18}{35}\)
b: \(=\dfrac{2}{5}\cdot\dfrac{-15}{8}=\dfrac{-30}{40}=-\dfrac{3}{4}\)
c: \(=\dfrac{2}{4}\cdot\dfrac{7}{3}=\dfrac{1}{2}\cdot\dfrac{7}{3}=\dfrac{7}{6}\)
d: \(=\dfrac{8}{3}\cdot\dfrac{16}{4}=\dfrac{128}{12}=\dfrac{32}{3}\)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`
a: \(=\dfrac{-7}{8}\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+3+\dfrac{7}{8}=\dfrac{-7}{8}+\dfrac{7}{8}+3=3\)
b: \(=-\dfrac{8}{5}:\dfrac{5}{3}=-\dfrac{24}{25}\)
c: \(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{6}{8}=\dfrac{6}{7}-\dfrac{5}{8}=\dfrac{48}{56}-\dfrac{35}{56}=\dfrac{13}{56}\)
A = 2 + 22 + 23 + ... + 28
2A = 22 + 23 + 24 + ... + 29
Lấy 2A trừ A theo vế ta có :
2A - A = (22 + 23 + 24 + ... + 29) - (2 + 22 + 23 + ... + 28)
A = 29 - 2
Vậy A = 29 - 2