— S = 1/4 + 2/4 +...+10/4 (1)
= 1 + 1/4 + 2/4 +...+ 9/4 (2)
=> Lấy (2) trừ đi (1) ta được:
1 — 10/4 = —6/4
Vì 14 = 14/1 = 84/6 mà —6/4 < 84/6
Do đó S < 14
— S = 1/4 + 2/4 +...+10/4 (1)
= 1 + 1/4 + 2/4 +...+ 9/4 (2)
=> Lấy (2) trừ đi (1) ta được:
1 — 10/4 = —6/4
Vì 14 = 14/1 = 84/6 mà —6/4 < 84/6
Do đó S < 14
1, Tính
\(-2\dfrac{1}{4}.\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right):\left(-2\dfrac{1}{8}\right)\)
2, Tính nhanh
\(\dfrac{4}{9}.19\dfrac{1}{3}-\dfrac{4}{9}.39\dfrac{1}{3}\) | \(19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}.\dfrac{12}{7}\)
\(\dfrac{-5}{7}.\dfrac{5}{11}+\dfrac{-5}{7}.\dfrac{2}{11}-\dfrac{5}{7}:\dfrac{11}{14}\) | \(\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4}{5}.\dfrac{82}{7}\)
\(\dfrac{5}{7}.\dfrac{-4}{19}+\dfrac{-15}{7}.\dfrac{5}{19}\) | \(8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
- Giải hộ với ạ, mấy anh, chị lớp 7,8 cũng được huhu :(( sáng mai nộp đề cương rồi. Làm ơn đi a
Muốn gì cũng hậu tạ :>
1. Tính giá trị biểu thức:
M=\(1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
2.Chứng tỏ rằng :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
3.So sánh
a, A=\(\dfrac{11^5+1}{11^6+1}\) ; B=\(\dfrac{11^6+1}{11^7+1}\)
b, M=\(\dfrac{15^{10}-1}{15^9+1}\) ; N=\(\dfrac{15^9-1}{15^{10}+1}\)
Tính giá trị biểu thức :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}\right)-\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{6}{7}\right)+...+\left(100+...+\dfrac{99}{100}\right)\)
Cho S = \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\). Hãy so sánh S và \(\dfrac{1}{2}\)
So sánh A vs \(\dfrac{3}{4}\)
Cho A= \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......+\dfrac{1}{200^2}\)
So sánh
\(A=\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{100}{2^{201}}\)
và \(B=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{100^3}\)
cho B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
so sánh B với 1
Cho \(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+................+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\) và \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+..........+\dfrac{48}{2}+\dfrac{49}{1}\)
Tính \(\dfrac{S}{P}\)
Help me!!!!!!!!!!!
BT2: Tính nhanh
5) \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\)
6) \(\dfrac{-3}{7}.\dfrac{19}{11}+\dfrac{-2}{7}.\dfrac{3}{11}+4\dfrac{6}{11}\)