Tính tổng sau : ( Dấu . là dấu nhân nhé )
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{999.1000}+1\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
dấu . là dấu nhân nha theo thcs
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
A = \(1-\frac{1}{100}=\frac{99}{100}\)
Bài 1 : tính tổng sau
P = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
S = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
Q = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
Dấu " . " là dấu nhân nhé
Bài 2 Tính tích P = ( \(1+\frac{1}{2}\)) X ( \(1+\frac{1}{3}\)) x ( \(1+\frac{1}{4}\)) x ... x ( \(1+\frac{1}{99}\))
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
1) Tính:
N = \(\frac{2}{2.5}\)+ \(\frac{2}{5.8}\)+...+ \(\frac{2}{17.20}\)
2) Tính hợp lý:
C = 1.2 +2.3 +3.4 +...+ 99.100
* Lưu ý: Dấu "." trong câu trên là dấu nhân nhé!
BÀI 1:
\(N=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{17.20}\)
\(N=2.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{17.20}\right)\)
\(N=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(N=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(N=2.\frac{9}{20}\)
\(N=\frac{9}{10}\)
BÀI 2:
\(C=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(3B=1.2\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(3B=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)
\(3B=99.100.101\)
\(3B=999900\)
\(\Rightarrow B=999900:3\)
\(B=333300\)
CHÚC BN HỌC TỐT!!!!
E=\(\frac{1}{1.2}\)-\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3}\)-\(\frac{1}{2.3.4}\)+\(\frac{1}{3.4}\)-\(\frac{1}{3.4.5}\)+...+\(\frac{1}{99.100}\)-\(\frac{1}{99.100.101}\)
(Dấu trừ viết hơi nhỏ và phần mẫu là dấu nhân nhé)
Tìm x, biết : 1.2 + 2.3 + 3.4 + ...... + 99.100 = \(2\frac{1}{5}x\) - 1
. là dấu nhân .
Tính tổng sau : A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
= \(1-\frac{1}{2017}\)
= \(\frac{2016}{2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{2016}+\frac{1}{2016}\right)-\frac{1}{2017}\)
\(A=1+0+0+...+0-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2017}{2017}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
Vậy: \(A=\frac{2016}{2017}\)
Cách làm của bạn Sang đầy đủ và chi tiết hơn đó bạn! :) Những bài có quy luật tương tự bạn cũng áp dụng cách giải trên nhé bạn.
Tính: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+..........+\frac{1}{997.998}+\frac{1}{999.1000}\)
Đặt Q = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{997.998}+\frac{1}{999.1000}\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{997.999}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{997}-\frac{1}{999}\)
\(2A=1-\frac{1}{999}\)
\(2A=\frac{998}{999}\)
\(\Leftrightarrow A=\frac{499}{999}\)
Đặt B = \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{998.1000}\)
\(2B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{998}-\frac{1}{1000}\)
\(2B=\frac{1}{2}-\frac{1}{1000}\)
\(B=\frac{499}{1000}\)
Vậy Q = A + B = \(\frac{499}{999}+\frac{499}{1000}\)
Tính: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{997.998}+\frac{1}{999.1000}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{999.1000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}=\frac{999}{1000}\)
Đặt
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{999.1000}\)
\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(\Leftrightarrow A=1-\frac{1}{1000}\)
\(\Leftrightarrow A=\frac{999}{1000}\)
Đặt:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow A=1-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{99}{100}\)
Tính: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{997.998}+\frac{1}{999.1000}\)
Đặt biểu thức là A.
A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
A=\(\frac{1}{1}-\frac{1}{1000}\)
A=\(\frac{999}{1000}\)