Question

BT1: Tính

3) \(\frac{7}{15}\) + \(\frac{9}{10}\)+ \(\frac{8}{15}\) + \(\frac{1}{10}\) + \(\frac{-20}{10}\) + \(\frac{1}{157}\)

Answer 1

Ta có: \(\dfrac{7}{15}+\dfrac{9}{10}+\dfrac{8}{15}+\dfrac{1}{10}+\dfrac{-20}{10}+\dfrac{1}{157}\)

\(=\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{9}{10}+\dfrac{1}{10}+\dfrac{-20}{10}\right)+\dfrac{1}{157}\)

\(=1+\left(-1\right)+\dfrac{1}{157}\)

\(=0+\dfrac{1}{157}=\dfrac{1}{157}.\)

Answer 2

Ta có: \(\dfrac{7}{15}+\dfrac{9}{10}+\dfrac{8}{15}+\dfrac{1}{10}+\dfrac{-20}{10}+\dfrac{1}{157}\)

= \(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{9}{10}+\dfrac{1}{10}+\dfrac{-20}{10}\right)+\dfrac{1}{157}\)

= \(\dfrac{7+8}{15}+\dfrac{10-20}{10}+\dfrac{1}{157}\)

= 1-1+\(\dfrac{1}{157}\)= 0 + \(\dfrac{1}{157}\) = \(\dfrac{1}{157}\)

Answer 3

\(\dfrac{7}{15}+\dfrac{9}{10}\)\(+\dfrac{8}{15}+\dfrac{1}{10}\)\(+\dfrac{-20}{10}+\dfrac{1}{157}\)

= \(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{9}{10}+\dfrac{1}{10}+\dfrac{-20}{10}\right)+\dfrac{1}{157}\)

= \(1+\left(-1\right)+\dfrac{1}{157}\)

= \(0+\dfrac{1}{157}\)

= \(\dfrac{1}{157}\)