Câu hỏi của Dương Thanh Ngân

Question

THỰC HIỆN PHÉP TÍNH SAU:

a/$\dfrac{1}{1.2}$+$\dfrac{1}{2.3}$+$\dfrac{1}{3.4}$+...........+$\dfrac{1}{99.100}$

b/$\dfrac{5.4^{15}-9^9-4.3^{10}.8^9}{5.29-6^{19}.7.2^{29}.27^6}$

Nguyễn Lan Anh · Accepted Answer

a) $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}$

= $\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}$

=$\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}$

=$1-\dfrac{1}{100}$

= $\dfrac{99}{100}$Câu trả lời: a) $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}$

= $\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}$

=$\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}$

=$1-\dfrac{1}{100}$

= $\dfrac{99}{100}$

Võ Đình Nhật · Answer

a) 11.2+12.3+13.4+....+199.10011.2+12.3+13.4+....+199.100

= 11−12+12−13+13−14+....+199−110011−12+12−13+13−14+....+199−1100

=11+0+0+...+0−110011+0+0+...+0−1100

=1−11001−1100

= 99100Câu trả lời: a) 11.2+12.3+13.4+....+199.10011.2+12.3+13.4+....+199.100

= 11−12+12−13+13−14+....+199−110011−12+12−13+13−14+....+199−1100

=11+0+0+...+0−110011+0+0+...+0−1100

=1−11001−1100

= 99100