Toán Tính nhanh : P=1/1×2+1/2×3+1/3×4+…..+1/2018×2019 12/09/2021 By aihong Tính nhanh : P=1/1×2+1/2×3+1/3×4+…..+1/2018×2019
$P = \dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + …. + \dfrac{1}{2018 \times 2019}$ $P =\dfrac{1}{1} – \dfrac{1}{2} + \dfrac{1}{2} -\dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{2018} -\dfrac{1}{2019}$ $P = 1 – \dfrac{1}{2019}$ $P = \dfrac{2018}{2019}$ Trả lời
Đáp án: $P=\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{2018×2019}$ $⇔P=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{2018}-\frac{1}{2019}$ $⇔P=1-\frac{1}{2019}=\frac{2018}{2019}$ #NOCOPY Trả lời
$P = \dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + …. + \dfrac{1}{2018 \times 2019}$
$P =\dfrac{1}{1} – \dfrac{1}{2} + \dfrac{1}{2} -\dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{2018} -\dfrac{1}{2019}$
$P = 1 – \dfrac{1}{2019}$
$P = \dfrac{2018}{2019}$
Đáp án:
$P=\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{2018×2019}$
$⇔P=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{2018}-\frac{1}{2019}$
$⇔P=1-\frac{1}{2019}=\frac{2018}{2019}$
#NOCOPY