Câu hỏi của Hảo Nguyễn

Question

Tính nhanh: $\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}$

Doraemon · Accepted Answer

Đặt $A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{98}+\dfrac{2}{192}$

$2A=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{98}+\dfrac{2}{192}$

$2A-A=\left(\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{98}+\dfrac{2}{192}\right)-\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{98}+\dfrac{2}{192}\right)$

$A=\dfrac{4}{3}-\dfrac{2}{192}$

$A=\dfrac{127}{96}$

Vậy $A=\dfrac{127}{96}$Câu trả lời: Đặt $A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{98}+\dfrac{2}{192}$

$2A=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{98}+\dfrac{2}{192}$

$2A-A=\left(\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{98}+\dfrac{2}{192}\right)-\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{98}+\dfrac{2}{192}\right)$

$A=\dfrac{4}{3}-\dfrac{2}{192}$

$A=\dfrac{127}{96}$

Vậy $A=\dfrac{127}{96}$

Doraemon · Answer

Mình hơn nhầm ở chỗ 98 viết lại thành 96. thông cảm nha mình làm lại cho:

Đặt $A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}$

$2A=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{96}+\dfrac{2}{192}$

$2A-A=\left(\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{96}+\dfrac{2}{192}\right)-\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\right)$

$A=\dfrac{4}{3}-\dfrac{2}{192}$

$A=\dfrac{127}{96}$

Vậy $A=\dfrac{127}{96}$Câu trả lời: Mình hơn nhầm ở chỗ 98 viết lại thành 96. thông cảm nha mình làm lại cho:

Đặt $A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}$

$2A=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{96}+\dfrac{2}{192}$

$2A-A=\left(\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{96}+\dfrac{2}{192}\right)-\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\right)$

$A=\dfrac{4}{3}-\dfrac{2}{192}$

$A=\dfrac{127}{96}$

Vậy $A=\dfrac{127}{96}$

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